lowest common ancestor
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Advanced Algorithms – COMS31900
Lowest Common Ancestor(with a bit on on Range Minimum Queries)
Benjamin Sach
Advanced Algorithms – COMS31900
Benjamin Sach
Lowest Common Ancestor(with a bit on on Range Minimum Queries)
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
After preprocessing,
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
After preprocessing,
? ! !? ?! ?
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
i
root
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
i
root
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
i
ancestors of node iroot
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
i
ancestors of node i
- nodes on the path
from i to the root
root
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
j
ancestors of node j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
j
ancestors of node j
- nodes on the path
from j to the root
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
i
j
root
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
rootcommon ancestors of i and j
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
rootcommon ancestors of i and j
- nodes which are
ancestors of both i
and j
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
root
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
root
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
4
5
3
0
1
2
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
root
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
depth =
4
5
3
0
1
2
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
root
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
depth =lowest common ancestor of i and j
4
5
3
0
1
2
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
i
j
root depth =lowest common ancestor of i and j
- the common ancestor of i
and j furthest
from the root
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
4
5
3
0
1
2
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
4
5
3
0
1
2
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
i
j 4
5
3
0
1
2LCA(i, j)
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
4
5
3
0
1
2
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
4
5
3
0
1
2
ij
LCA(i, j)
Lowest common ancestor
Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .
After preprocessing,
the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j
4
5
3
0
1
2
ij
LCA(i, j)
• Ideally, we would like O(n) space, O(n) prep. time and O(1) query time
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
the nodes are numbered between
0 and (n− 1)
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
Write down every node you visit. . . and its depth
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
0(node) N
Write down every node you visit. . . and its depth
(depth) D 0
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
10(node) N
Write down every node you visit. . . and its depth
(depth) D 10
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 50(node) N
Write down every node you visit. . . and its depth
(depth) D 210
7
0
9
6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 50(node) N
Write down every node you visit. . . and its depth
(depth) D 3210
7
0
9
6
9
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 50(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2
7
0
9
6
9
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 100(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3
7
0
9
6
9
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 50(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2
7
0
9
6
9
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 10(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1
7
0
9
6
9
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 10(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2
7
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 10(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1
7
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 00(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 20(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 00(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 30(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 30(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 30(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 80(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 30(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 00(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 40(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
how do we find LCA(i,j)?
7
i j
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
how do we find LCA(i,j)?
7
i j
Find i and j in N
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
how do we find LCA(i,j)?
7
i j
Find i and j in N
7
i′ j′0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
how do we find LCA(i,j)?
7
i j
Find i and j in N
7
i′ j′
Compute RMQ(i′, j′) in D
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
how do we find LCA(i,j)?
7
i j
Find i and j in N
7
i′ j′
Compute RMQ(i′, j′) in D
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
ji
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
ji
9
6
Find i and j in N
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
ji
9
6
Find i and j in N
i′ j′
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
ji
9
6
Find i and j in N
i′ j′
Compute RMQ(i′, j′) in D
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
ji
9
6
Find i and j in N
i′ j′
Compute RMQ(i′, j′) in D
9 6
1
11
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
9
6
9 6
1
ji
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
9
6
9 6
1
ji
Find i and j in N . . .
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
9
6
9 6
1
ji
Find i and j in N . . . which copy of i?
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
Compute an Euler tour of T . . .(a depth first search with repeats)
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
How long is the tour?
We follow each edge twice. . .and there are (n− 1) edges
Write down every node you visit. . . and its depth
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
how do we find LCA(i,j)?
9
6
9 6
1
ji
Find i and j in N . . . which copy of i? any copy is fine
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(n)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(n)
O(?)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
O(1)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
O(1)
O(?)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
O(1)
O(?)
O(1)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
O(1)
O(?)
O(1)
Prep. time O(n+ prepRMQ(n))
Query time O(1 + queryRMQ(n))
Space O(n+ spaceRMQ(n))
depends on the RMQ structure used
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
O(1)
O(?)
O(1)
Prep. time O(n log logn)
Query time O(1)
Space O(n log logn)
using the best result from last lecture
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
why does this work?
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(?)
O(1)
O(?)
O(1)
Prep. time O(n log logn)
Query time O(1)
Space O(n log logn)
using the best result from last lecture
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
depth > d+ 1
S2
x
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
children
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
depth > d+ 1
S2
x
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
subtrees
children
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
depth > d+ 1
S2
x
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
subtrees
children
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
S2
x
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
S2
x
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i j
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i j imagine LCA(i, j) is not y
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i j imagine LCA(i, j) is not y
i′ and j′ are in here so RMQ does not return the location of a y
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i j imagine LCA(i, j) is not y
i′ and j′ are in here so RMQ does not return the location of a y
x
y
(all of the ys are out of range)
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i
j
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i
j
again, imagine LCA(i, j) is not y
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
i
j
again, imagine LCA(i, j) is not y
i′ and j′ cross an x (which has smaller depth than y)
so the RMQ location isn’t a y
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
i j
y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
i j
y
now imagine LCA(i, j) is y
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
i j
y
now imagine LCA(i, j) is y
i′ and j′ cross a y (which is the smallest in the range)
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
i j
y
now imagine LCA(i, j) is y
i′ and j′ cross a y (which is the smallest in the range)
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
i j
y
now imagine LCA(i, j) is y
i′ and j′ cross a y (which is the smallest in the range)
Solving LCA using RMQ - correctness
We can also define a Euler tour of T recursively. . .
S1 S3 Sk
depth = d
depth = d+ 1
x x xx(node) N
(depth) D
x x
d d dd d d
tour of S1 tour of S2 tour of S3 tour of Sk
depth > d+ 1
Claim the RMQ reports the location of
some y in N iff LCA(i, j) = y
y yy y
d+ 1 d+ 1d+ 1 d+ 1
x
i j
y
now imagine LCA(i, j) is y
i′ and j′ cross a y (which is the smallest in the range)so the RMQ reports a y
Ongoing Summary
We have seen an O(n log logn) space, O(n log logn) prep. time and O(1) query time solution
for the Lowest Common Ancestor problemwhich uses solution 3 for RMQ from last lecture
Ongoing Summary
Can we do better?
We have seen an O(n log logn) space, O(n log logn) prep. time and O(1) query time solution
for the Lowest Common Ancestor problemwhich uses solution 3 for RMQ from last lecture
Solving LCAs using RMQs - efficiency
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
Solving LCAs using RMQs - efficiency
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
Notice anything interesting about D?
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
Solving LCAs using RMQs - efficiency
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
Notice anything interesting about D? D[i+ 1] = D[i]± 1
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
±1 Range minimum query
A
Preprocess an integer array A (length n) to answer range minimum queries. . .
n
After preprocessing, a range minimum query is given by RMQ(i, j)
the output is the location of the smallest element in A[i, j]
e.g. RMQ(3, 7) = 5, which is the location of the smallest element in A[3, 7]
i = 3 j = 7
RMQ(3, 7) = 5
• Can we exploit this±1 property to get a more efficient RMQ data structure?
• Ideally we would like O(n) space, O(n) prep. time and O(1) query time
where for all k, we have A[k + 1] = A[k]± 1
16 1715 16 15 14 15 16 17 18 19 20 21 20 19 20
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(in a tie, report the leftmost)
Low-resolution RMQ (again)
A
Key Idea replace A with a smaller, ‘low resolution’ array H
H
n
i j
and many small arrays L0, L1, L2 . . . ‘for the details’
n
i′ j′
Preprocess the array H(
which has length n = 2nlogn
)to answer RMQs. . .
in O(n) space/prep time
Preprocess each array Li (which has length (logn)/2) to answer RMQs. . .
in O(logn log logn) space/prep time
How do we answer a query in A in O(1) time?
Do one query in H and one query in two different Li and return the smallest
n = 2nlogn
logn2
min of these
as there are O(n/ logn) Li arrays, we have O(n log logn) total space/prep time
L0L1
L2L3
L4L5 Ln
goes in here
all of thesego in here
Low-resolution RMQ (again)
A
Key Idea replace A with a smaller, ‘low resolution’ array H
H
n
i j
and many small arrays L0, L1, L2 . . . ‘for the details’
n
i′ j′
Preprocess the array H(
which has length n = 2nlogn
)to answer RMQs. . .
in O(n) space/prep time
Preprocess each array Li (which has length (logn)/2) to answer RMQs. . .
in O(logn log logn) space/prep time
How do we answer a query in A in O(1) time?
Do one query in H and one query in two different Li and return the smallest
n = 2nlogn
logn2
min of these
as there are O(n/ logn) Li arrays, we have O(n log logn) total space/prep time
too big and slow!
L0L1
L2L3
L4L5 Ln
goes in here
all of thesego in here
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4RMQx(0, 2) = RMQy(0, 2) = 2
RMQx(3, 4) = RMQy(3, 4) = 4
RMQx(0, 4) = RMQy(0, 4) = 2
RMQx(0, 1) = RMQy(0, 1) = 1
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4
0100 0100
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4
0100 0100
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4
0100 0100dx = dy == 2 = 2
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
iff dx = dy
0100 0100dx = dy == 2 = 2
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
iff dx = dy
0100 0100dx = dy == 2 = 2
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
iff dx = dy
0100 0100dx = dy == 2 = 2
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
iff dx = dy
0100 0100dx = dy == 2 = 2
d contains (logn)/2− 1 bits so . . .
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
iff dx = dy
0100 0100dx = dy == 2 = 2
d contains (logn)/2− 1 bits so . . . at most 2(logn)/2
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
iff dx = dy
0100 0100dx = dy == 2 = 2
d contains (logn)/2− 1 bits so . . . at most 2(logn)/2 =(2logn
)1/2
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
iff dx = dy
0100 0100dx = dy == 2 = 2
d contains (logn)/2− 1 bits so . . . at most 2(logn)/2 =(2logn
)1/26√n
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
• For each value of d we store RMQ(i, j) for all i, j
iff dx = dy
0100 0100dx = dy == 2 = 2
d contains (logn)/2− 1 bits so . . . at most 2(logn)/2 =(2logn
)1/26√n
(remember these are the locations of the minimum)
Counting ±1 RMQ arrays
L
logn2
How many different±1 RMQ arrays like this. . . are there?
Ly
logn2
We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)
Lx
logn2
is equivalent to
16 15 14 15 14 10 9 8 9 8is equivalent to
Lx Ly0 1 2 3 4 0 1 2 3 4
16 15 14 15 14 10 9 8 9 8
0 1 2 3 4 0 1 2 3 4 Fact Lx is equivalent to Ly
• We can precompute dx for each Lx in O(|Lx|) = O(logn) time.
• How many different values of d are there?
• For each value of d we store RMQ(i, j) for all i, j
iff dx = dy
. . . this requires O(√n log2 n) = O(n) total space and prep. time
0100 0100dx = dy == 2 = 2
d contains (logn)/2− 1 bits so . . . at most 2(logn)/2 =(2logn
)1/26√n
(remember these are the locations of the minimum)
RMQ on the L arrays in linear space
A
Key Idea replace A with a smaller, ‘low resolution’ array H
n
n = 2nlogn
logn2
L0L1
L2L3
L4L5 Ln
d0 d1 d2 d3 d4 d5 dn
precompute the value of dx for each Lx
in O(n) total space and prep. time
Precompute all the RMQ answers for
every value 0 6 d 6√n
in O(n) total space and prep. time
To perform a query within some Lx
• Look up dx
• Find the row dx in the table
• Find the entry giving RMQx(i, j)
row d3
This takes O(1) time
Optimal ±1 RMQKey Idea replace A with a smaller, ‘low resolution’ array H
and many small arrays L0, L1, L2 . . . ‘for the details’
Preprocess the array H (which has length n = 8nlogn ) to answer RMQs. . .
in O(n) space/prep time
Preprocess each array Li (which has length (logn)/2) to answer RMQs. . .
in O(logn) space/prep time
How do we answer a query in A in O(1) time?
Do one query in H and one query in two different Li and return the smallest
n = 2nlogn
as there are O(n/ logn) Li arrays, we have O(n) total space/prep time
A
H
n
i jn
i′ j′
logn2
min of these
L0L1
L2L3
L4L5 Ln
goes in here
all of thesego in here
Ongoing Summary
for the Lowest Common Ancestor problem
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the ±1 Range Minimum Query problemwhich improves solution 3 for RMQ from last lecture
(but only for±1 inputs)
which uses solution 3 for RMQ from last lecture
We have seen an O(n log logn) space, O(n log logn) prep. time and O(1) query time solution
Ongoing Summary
How does this affect our LCA solution?
for the Lowest Common Ancestor problem
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the ±1 Range Minimum Query problemwhich improves solution 3 for RMQ from last lecture
(but only for±1 inputs)
which uses solution 3 for RMQ from last lecture
We have seen an O(n log logn) space, O(n log logn) prep. time and O(1) query time solution
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(n)
O(1)
O(1)
O(1)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(n)
O(1)
O(1)
O(1)
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(n)
O(1)
O(1)
O(1)
This gives us O(n) space, O(n) prep. time and O(1) query time for the LCA problem
Solving LCAs using RMQs
3
0
1
2
2 3 4
5 8
10
1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 00(node) N
(depth) D 3210 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0
2n−1
7
7
0
0
9
6
9 6
1
Preprocessing Summary
1. Construct N and D from T
2. Add a pointer from each
3. Preprocess D for RMQs
node i to some N [i′] = i
Query Summary - LCA(i,j)
1. Find (any) i′ st. N [i′] = i
2. Find (any) j′ st. N [j′] = j
3. Compute RMQ(i′, j′) in D
4. LCA(i, j) = N [RMQ(i′, j′)]
O(n)
O(1)
O(n)
O(n)
O(1)
O(1)
O(1)
This gives us O(n) space, O(n) prep. time and O(1) query time for the LCA problemby using the solution to ±1RMQ
Ongoing Summary
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the Lowest Common Ancestor problemwhich uses the solution to ±1RMQ
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the ±1 Range Minimum Query problemwhich improves solution 3 for RMQ from last lecture
(but only for±1 inputs)
Ongoing Summary
What about the general Range Minimum Query problem?
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the Lowest Common Ancestor problemwhich uses the solution to ±1RMQ
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the ±1 Range Minimum Query problemwhich improves solution 3 for RMQ from last lecture
(but only for±1 inputs)
(when the inputs aren’t±1)
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
Build the Cartesian tree, TA of the array A:
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected locationpartitions the array in two
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
17 823 73 51 82 19 32 67 91 14 46 9 21 54
partitions the array in two
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
17 823 73 51 82 19 32 67 91 14 46 9 21 54
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
17 823 73 51 82 19 32 67 91 14 46 9 21 54
8 9
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location 8 9
1723 73 51 82 19 32 67 91 14 46 21 54
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6 13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location 8 9
1723 73 51 82 19 32 67 91 14 46 21 54
17 14 21
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location 8 9
17 14 21
23 73 51 82 32 67 91 46 54
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location 8 9
17 14 21
23 73 51 82 32 67 91 46 54
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location 8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location 8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
Key Fact: The LCA in TA equals the RMQ in A
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46
i = 3 j = 7
RMQ(3, 7) = 6
9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
Key Fact: The LCA in TA equals the RMQ in A
6
23 51 32 67 46 54
73
32
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
Key Fact: The LCA in TA equals the RMQ in A
6
23 51 67 46 5432
19
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
Key Fact: The LCA in TA equals the RMQ in A
6
23 51 67 46 5432
19
9
91
54
i = 10 j = 15
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
Key Fact: The LCA in TA equals the RMQ in A
This gives us O(n) space, O(n) prep. time and O(1) query time for the RMQ problem
6
23 51 67 46 5432
19
9
91
54
i = 10 j = 15
13
Solving RMQs using LCAs
A
n
17 823 73 51 82 19 32 5 67 91 14 46 9 21 54
0 1 2 3 4 5 7 8 9 10 11 12 14 15
5
Build the Cartesian tree, TA of the array A:
• The root is the smallest value
• The selected location
This process isn’t very efficient. . .
a better one takes O(n) time
8 9
17 14 21
73 82 91
partitions the array in two
• The rest of the tree is given byrecursing left and right. . .
it’s not tricky but we don’t have
time to cover it
Key Fact: The LCA in TA equals the RMQ in A
This gives us O(n) space, O(n) prep. time and O(1) query time for the RMQ problem
6
23 51 67 46 5432
19
9
91
54
i = 10 j = 15
13
by using the solution to LCA :)
Summary
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the Range Minimum Query problemwhich uses the solution to LCA
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the Lowest Common Ancestor problemwhich uses the solution to ±1RMQ
We have seen an O(n) space, O(n) prep. time and O(1) query time solution
for the ±1 Range Minimum Query problemwhich improves solution 3 for RMQ from last lecture
(but only for±1 inputs)
(which works for all inputs)
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