6/10/2015 DynamicProgramming|Set12(LongestPalindromicSubsequence)GeeksforGeeks

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DynamicProgramming|Set12(LongestPalindromicSubsequence)

Givenasequence,findthelengthofthelongestpalindromicsubsequenceinit.Forexample,ifthegivensequenceisBBABCBCAB,thentheoutputshouldbe7asBABCBABisthelongestpalindromicsubseuqnceinit.BBBBBandBBCBBarealsopalindromicsubsequencesofthegivensequence,butnotthelongestones.

Thenaivesolutionforthisproblemistogenerateallsubsequencesofthegivensequenceandfindthelongestpalindromicsubsequence.Thissolutionisexponentialintermoftimecomplexity.LetusseehowthisproblempossessesbothimportantpropertiesofaDynamicProgramming(DP)ProblemandcanefficientlysolvedusingDynamicProgramming.

1)OptimalSubstructure:LetX[0..n1]betheinputsequenceoflengthnandL(0,n1)bethelengthofthelongestpalindromicsubsequenceofX[0..n1].

IflastandfirstcharactersofXaresame,thenL(0,n1)=L(1,n2)+2.ElseL(0,n1)=MAX(L(1,n1),L(0,n2)).

Followingisageneralrecursivesolutionwithallcaseshandled.

//Everaysinglecharacterisapalindromoflength1

L(i,i)=1forallindexesiingivensequence

//IFfirstandlastcharactersarenotsame

If(X[i]!=X[j])L(i,j)=max{L(i+1,j),L(i,j1)}

//Ifthereareonly2charactersandbotharesame

Elseif(j==i+1)L(i,j)=2

//Iftherearemorethantwocharacters,andfirstandlast

//charactersaresame

ElseL(i,j)=L(i+1,j1)+2

2)OverlappingSubproblemsFollowingissimplerecursiveimplementationoftheLPSproblem.Theimplementationsimplyfollowstherecursivestructurementionedabove.

#include#include

6/10/2015 DynamicProgramming|Set12(LongestPalindromicSubsequence)GeeksforGeeks

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Output:

ThelnegthoftheLPSis5

Consideringtheaboveimplementation,followingisapartialrecursiontreeforasequenceoflength6withalldifferentcharacters.

L(0,5)

/\

/\

L(1,5)L(0,4)

/\/\

/\/\

L(2,5)L(1,4)L(1,4)L(0,3)

Intheabovepartialrecursiontree,L(1,4)isbeingsolvedtwice.Ifwedrawthecompleterecursiontree,thenwecanseethattherearemanysubproblemswhicharesolvedagainandagain.Sincesamesuproblemsarecalledagain,thisproblemhasOverlappingSubprolemsproperty.SoLPSproblemhasboth

//Autilityfunctiontogetmaxoftwointegersintmax(intx,inty){return(x>y)?x:y;}//Returnsthelengthofthelongestpalindromicsubsequenceinseqintlps(char*seq,inti,intj){//BaseCase1:Ifthereisonly1characterif(i==j)return1;//BaseCase2:Ifthereareonly2charactersandbotharesameif(seq[i]==seq[j]&&i+1==j)return2;//Ifthefirstandlastcharactersmatchif(seq[i]==seq[j])returnlps(seq,i+1,j1)+2;//Ifthefirstandlastcharactersdonotmatchreturnmax(lps(seq,i,j1),lps(seq,i+1,j));}/*Driverprogramtotestabovefunctions*/intmain(){charseq[]="GEEKSFORGEEKS";intn=strlen(seq);printf("ThelnegthoftheLPSis%d",lps(seq,0,n1));getchar();return0;}

6/10/2015 DynamicProgramming|Set12(LongestPalindromicSubsequence)GeeksforGeeks

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properties(seethisandthis)ofadynamicprogrammingproblem.LikeothertypicalDynamicProgramming(DP)problems,recomputationsofsamesubproblemscanbeavoidedbyconstructingatemporaryarrayL[][]inbottomupmanner.

DynamicProgrammingSolution

#include#include//Autilityfunctiontogetmaxoftwointegersintmax(intx,inty){return(x>y)?x:y;}//Returnsthelengthofthelongestpalindromicsubsequenceinseqintlps(char*str){intn=strlen(str);inti,j,cl;intL[n][n];//Createatabletostoreresultsofsubproblems//Stringsoflength1arepalindromeoflentgh1for(i=0;i

6/10/2015 DynamicProgramming|Set12(LongestPalindromicSubsequence)GeeksforGeeks

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Output:

ThelnegthoftheLPSis7

TimeComplexityoftheaboveimplementationisO(n^2)whichismuchbetterthantheworstcasetimecomplexityofNaiveRecursiveimplementation.

ThisproblemisclosetotheLongestCommonSubsequence(LCS)problem.Infact,wecanuseLCSasasubroutinetosolvethisproblem.FollowingisthetwostepsolutionthatusesLCS.1)Reversethegivensequenceandstorethereverseinanotherarraysayrev[0..n1]2)LCSofthegivensequenceandrev[]willbethelongestpalindromicsequence.ThissolutionisalsoaO(n^2)solution.

Pleasewritecommentsifyoufindanythingincorrect,oryouwanttosharemoreinformationaboutthetopicdiscussedabove.

References:http://users.eecs.northwestern.edu/~dda902/336/hw6sol.pdf

return0;}