Following steps build L [ m + 1 ] [ n + 1 ] in bottom up fashion . Note that L [ i ] [ j ] contains cost of removing identical characters in LCS of X [ 0. . i - 1 ] and Y [ 0. . j - 1 ]
for ( int i = 0 ; i <= m ; ++ i ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 j == 0 ) L [ i ] [ j ] = 0 ;