text: Longest Common Subsequence | DP | Dynamic Programming C implementation of LCS problem ; Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ] ; Following steps build L [ m + 1 ] [ n + 1 ] in bottom up fashion . Note that L [ i ] [ j ] contains length of LCS of X [ 0. . i - 1 ] and Y [ 0. . j - 1 ] ; L [ m ] [ n ] contains length of LCS for X [ 0. . n - 1 ] and Y [ 0. . m - 1 ] ; Utility function to get max of 2 integers ; Driver program to test above function
code: #include <bits/stdc++.h> NEW_LINE int max ( int a , int b ) ; int lcs ( char * X , char * Y , int m , int n ) { int L [ m + 1 ] [ n + 1 ] ; int i , j ; for ( i = 0 ; i <= m ; i ++ ) { for ( j = 0 ; j <= n ; j ++ ) { if ( i == 0 j == 0 ) L [ i ] [ j ] = 0 ; else if ( X [ i - 1 ] == Y [ j - 1 ] ) L [ i ] [ j ] = L [ i - 1 ] [ j - 1 ] + 1 ; else L [ i ] [ j ] = max ( L [ i - 1 ] [ j ] , L [ i ] [ j - 1 ] ) ; } } return L [ m ] [ n ] ; } int max ( int a , int b ) { return ( a > b ) ? a : b ; } int main ( ) { char X [ ] = " AGGTAB " ; char Y [ ] = " GXTXAYB " ; int m = strlen ( X ) ; int n = strlen ( Y ) ; printf ( " Length ▁ of ▁ LCS ▁ is ▁ % d " , lcs ( X , Y , m , n ) ) ; return 0 ; }