Longest common sub-sequence problem (LCS)

Given two strings s and T. Find the length of the longest common subsequence of these two strings.

Input:

N=4

M=4

s= "ABCD"

t= "BECD"

Output:

3 ("BCD")

This type of problem is known as the longest common sub-sequence problem (Lcs,longest Common subsequence).

Max (Dp[i][j]+1,dp[i][j+1],dp[i+1][j]) (s=t)

dp[i+1][j+1]=

Max (Dp[i][j+1],dp[i+1][j]) (other)

This recursion can be calculated using O (nm), dp[n][m] is the length of the LCS.

J\i	0	1{B}	2{e}	3{C}	4{D}
0	0	0	0	0	0
1{a}	0	0	0	0	0
2{B}	0	1	1	1	1
3{C}	0	1	1	2	2
4{D}	0	1	1	2	3

1 intn,m;2 CharS[max],t[max];3 intDp[max][max];//DP Array4 5 voidSolve ()6 {7      for(intI=0; i<=n; i++){8          for(intj=0; j<=m; J + +){9             if(s[i]==T[j]) {Tendp[i+1][j+1]=dp[i][j]+1; One             } A             Else{ -dp[i+1][j+1]=max (dp[i][j+1],dp[i+1][j]); -             } the         } -     } -printf"%d\n", Dp[n][w]); -}

From:<< Challenge Program Design Competition >>

Longest common sub-sequence problem (LCS)