Longest Common Subsequence | DP using Memoization | Java program to memoize recursive implementation of LCS problem ; Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ] memoization applied in recursive solution ; base case ; if the same state has already been computed ; if equal , then we store the value of the function call ; store it in arr to avoid further repetitive work in future function calls ; store it in arr to avoid further repetitive work in future function calls ; Driver Code ; assign - 1 to all positions
import java . util . Arrays ; class GFG { static final int maximum = 1000 ; static int lcs ( String X , String Y , int m , int n , int dp [ ] [ ] ) { if ( m == 0 n == 0 ) { return 0 ; } if ( dp [ m - 1 ] [ n - 1 ] != - 1 ) { return dp [ m - 1 ] [ n - 1 ] ; } if ( X . charAt ( m - 1 ) == Y . charAt ( n - 1 ) ) { dp [ m - 1 ] [ n - 1 ] = 1 + lcs ( X , Y , m - 1 , n - 1 , dp ) ; return dp [ m - 1 ] [ n - 1 ] ; } else { dp [ m - 1 ] [ n - 1 ] = Math . max ( lcs ( X , Y , m , n - 1 , dp ) , lcs ( X , Y , m - 1 , n , dp ) ) ; return dp [ m - 1 ] [ n - 1 ] ; } } public static void main ( String [ ] args ) { String X = " AGGTAB " ; String Y = " GXTXAYB " ; int m = X . length ( ) ; int n = Y . length ( ) ; int dp [ ] [ ] = new int [ m ] [ maximum ] ; for ( int [ ] row : dp ) { Arrays . fill ( row , - 1 ) ; } System . out . println ( " Length ▁ of ▁ LCS : ▁ " + lcs ( X , Y , m , n , dp ) ) ; } }