Minimum Cost To Make Two Strings Identical | Java code to find minimum cost to make two strings identical ; 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 ] ; Returns cost of making X [ ] and Y [ ] identical . costX is cost of removing a character from X [ ] and costY is cost of removing a character from Y [ ] / ; Find LCS of X [ ] and Y [ ] ; Cost of making two strings identical is SUM of following two 1 ) Cost of removing extra characters from first string 2 ) Cost of removing extra characters from second string ; Driver code
import java . io . * ; class GFG { static int lcs ( String X , String Y , int m , int n ) { int L [ ] [ ] = new int [ m + 1 ] [ n + 1 ] ; for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 j == 0 ) L [ i ] [ j ] = 0 ; else if ( X . charAt ( i - 1 ) == Y . charAt ( j - 1 ) ) L [ i ] [ j ] = L [ i - 1 ] [ j - 1 ] + 1 ; else L [ i ] [ j ] = Math . max ( L [ i - 1 ] [ j ] , L [ i ] [ j - 1 ] ) ; } } return L [ m ] [ n ] ; } static int findMinCost ( String X , String Y , int costX , int costY ) { int m = X . length ( ) ; int n = Y . length ( ) ; int len_LCS ; len_LCS = lcs ( X , Y , m , n ) ; return costX * ( m - len_LCS ) + costY * ( n - len_LCS ) ; } public static void main ( String [ ] args ) { String X = " ef " ; String Y = " gh " ; System . out . println ( " Minimum ▁ Cost ▁ to ▁ make ▁ two ▁ strings ▁ " + " ▁ identical ▁ is ▁ = ▁ " + findMinCost ( X , Y , 10 , 20 ) ) ; } }