Given two strings s1 and s2, the task is to find the length of the longest common subsequence present in both of them.
Examples:
Input: s1 = “ABCDGH”, s2 = “AEDFHR”
Output: 3
LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.Input: s1 = “striver”, s2 = “raj”
Output: 1
The naive solution for this problem is to generate all subsequences of both given sequences and find the longest matching subsequence. This solution is exponential in terms of time complexity. The general recursive solution of the problem is to generate all subsequences of both given sequences and find the longest matching subsequence. The total possible combinations will be 2n. Hence, the recursive solution will take O(2n).
Optimal Substructure:
- Let the input sequences are X[0… m-1] and Y[0…n-1] of lengths m and n respectively. And let L(X[0… m-1], Y[0…n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0… m-1], Y[0…n-1]).
- If last characters of both sequences match (or X[m-1] == Y[n-1]) then L(X[0… m-1], Y[0…n-1]) = 1 + L(X[0… m-2], Y[0…n-2])
- If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then L(X[0… m-1], Y[0…n-1]) = MAX (L(X[0… m-2], Y[0…n-1]), L(X[0… m-1], Y[0…n-2])
Given below is the recursive solution to the LCS problem:
Implementation:
C++
// A Naive C++ recursive implementation// of LCS of two strings#include <bits/stdc++.h>using namespace std;// Returns length of LCS for X[0..m-1], Y[0..n-1]int lcs(string X, string Y, int m, int n){ if (m == 0 || n == 0) return 0; if (X[m - 1] == Y[n - 1]) return 1 + lcs(X, Y, m - 1, n - 1); else return max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, n));}// Driver Codeint main(){ string X = "AGGTAB"; string Y = "GXTXAYB"; // Find the length of string int m = X.length(); int n = Y.length(); cout << "Length of LCS: " << lcs(X, Y, m, n); return 0;} |
Java
// A Naive Java recursive implementation // of LCS of two strings class GFG {// Returns length of LCS for X[0..m-1], Y[0..n-1] static int lcs(String X, String Y, int m, int n) { if (m == 0 || n == 0) { return 0; } if (X.charAt(m - 1) == Y.charAt(n - 1)) { return 1 + lcs(X, Y, m - 1, n - 1); } else { return Math.max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, n)); } }// Driver Code public static void main(String[] args) { String X = "AGGTAB"; String Y = "GXTXAYB"; // Find the length of String int m = X.length(); int n = Y.length(); System.out.println("Length of LCS: " + lcs(X, Y, m, n)); }}// This code is contributed by 29AjayKumar |
Python3
# A Naive Python recursive implementation# of LCS of two strings# Returns length of LCS for X[0..m-1], Y[0..n-1]def lcs(X, Y, m, n): if (m == 0 or n == 0): return 0 if (X[m - 1] == Y[n - 1]): return 1 + lcs(X, Y, m - 1, n - 1) else: return max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, n))# Driver Codeif __name__ == '__main__': X = "AGGTAB" Y = "GXTXAYB" # Find the length of string m = len(X) n = len(Y) print("Length of LCS:", lcs(X, Y, m, n))# This code is contributed by 29AjayKumar |
C#
// A Naive recursive C#implementation of // LCS of two strings using System;class GFG{// Returns length of LCS for // X[0..m-1], Y[0..n-1] static int lcs(String X, String Y, int m, int n) { if (m == 0 || n == 0) { return 0; } if (X[m - 1] == Y[n - 1]) { return 1 + lcs(X, Y, m - 1, n - 1); } else { return Math.Max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, n)); }}// Driver Code public static void Main() { String X = "AGGTAB"; String Y = "GXTXAYB"; // Find the length of String int m = X.Length; int n = Y.Length; Console.Write("Length of LCS: " + lcs(X, Y, m, n));}}// This code is contributed by 29AjayKumar |
PHP
<?php // A Naive PHP recursive implementation// of LCS of two strings// Returns length of LCS for// X[0..m-1], Y[0..n-1]function lcs($X, $Y, $m, $n){ if ($m == 0 || $n == 0) return 0; if ($X[$m - 1] == $Y[$n - 1]) return 1 + lcs($X, $Y, $m - 1, $n - 1); else return max(lcs($X, $Y, $m, $n - 1), lcs($X, $Y, $m - 1, $n));}// Driver Code$X = "AGGTAB";$Y = "GXTXAYB";// Find the length of string$m = strlen($X);$n = strlen($Y);echo "Length of LCS: " . lcs($X, $Y, $m, $n);// This code is contributed by ita_c?> |
Javascript
<script>// A Naive Javascript recursive implementation // of LCS of two strings // Returns length of LCS for X[0..m-1], Y[0..n-1] function lcs(X,Y,m,n) { if (m == 0 || n == 0) { return 0; } if (X[m-1] == Y[n-1]) { return 1 + lcs(X, Y, m - 1, n - 1); } else { return Math.max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, n)); } } // Driver Code let X = "AGGTAB"; let Y = "GXTXAYB"; // Find the length of String let m = X.length; let n = Y.length; document.write("Length of LCS: " + lcs(X, Y, m, n));// This code is contributed by rag2127</script> |
Output
Length of LCS: 4
Dynamic Programming using Memoization
Considering the above implementation, the following is a partial recursion tree for input strings “AXYT” and “AYZX”
lcs("AXYT", "AYZX")
/ \
lcs("AXY", "AYZX") lcs("AXYT", "AYZ")
/ \ / \
lcs("AX", "AYZX") lcs("AXY", "AYZ") lcs("AXY", "AYZ") lcs("AXYT", "AY")
In the above partial recursion tree, lcs(“AXY”, “AYZ”) are being solved twice. On drawing the complete recursion tree, it has been observed that there are many subproblems that are solved again and again. So this problem has Overlapping Substructure property and recomputation of same subproblems can be avoided by either using Memoization or Tabulation. The tabulation method has been discussed here.
A common point of observation to use memoization in the recursive code will be the two non-constant arguments M and N in every function call. The function has 4 arguments, but 2 arguments are constant which does not affect the Memoization. The repetitive calls occur for N and M which have been called previously. Following the below steps will help us to write the DP solution using memoization.
- Use a 2-D array to store the computed lcs(m, n) value at arr[m-1][n-1] as the string index starts from 0.
- Whenever the function with the same argument m and n are called again, do not perform any further recursive call and return arr[m-1][n-1] as the previous computation of the lcs(m, n) has already been stored in arr[m-1][n-1], hence reducing the recursive calls that happen more than once.
Below is the implementation of the above approach:
C++
// C++ program to memoize// recursive implementation of LCS problem#include <bits/stdc++.h>using namespace std;const int maximum = 1000;// Returns length of LCS for X[0..m-1], Y[0..n-1] */// memoization applied in recursive solutionint lcs(string X, string Y, int m, int n, int dp[][maximum]){ // base case if (m == 0 || n == 0) return 0; // if the same state has already been // computed if (dp[m - 1][n - 1] != -1) return dp[m - 1][n - 1]; // if equal, then we store the value of the // function call if (X[m - 1] == Y[n - 1]) { // store it in arr to avoid further repetitive // work in future function calls dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp); return dp[m - 1][n - 1]; } else { // store it in arr to avoid further repetitive // work in future function calls dp[m - 1][n - 1] = max(lcs(X, Y, m, n - 1, dp), lcs(X, Y, m - 1, n, dp)); return dp[m - 1][n - 1]; }}// Driver Codeint main(){ string X = "AGGTAB"; string Y = "GXTXAYB"; int m = X.length(); int n = Y.length(); int dp[m][maximum]; // assign -1 to all positions memset(dp, -1, sizeof(dp)); cout << "Length of LCS: " << lcs(X, Y, m, n, dp); return 0;} |
Java
import java.util.Arrays;// Java program to memoize// recursive implementation of LCS problem class GFG { static final int maximum = 1000;// Returns length of LCS for X[0..m-1], Y[0..n-1] */ // memoization applied in recursive solution static int lcs(String X, String Y, int m, int n, int dp[][]) { // base case if (m == 0 || n == 0) { return 0; } // if the same state has already been // computed if (dp[m - 1][n - 1] != -1) { return dp[m - 1][n - 1]; } // if equal, then we store the value of the // function call if (X.charAt(m - 1) == Y.charAt(n - 1)) { // store it in arr to avoid further repetitive // work in future function calls dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp); return dp[m - 1][n - 1]; } else { // store it in arr to avoid further repetitive // work in future function calls 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]; } }// Driver Code 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]; // assign -1 to all positions for (int[] row : dp) { Arrays.fill(row, -1); } System.out.println("Length of LCS: " + lcs(X, Y, m, n, dp)); }}/* This Java code is contributed by 29AjayKumar*/ |
Python3
# Python3 program to memoize# recursive implementation of LCS problemmaximum = 1000# Returns length of LCS for X[0..m-1], Y[0..n-1] */# memoization applied in recursive solutiondef lcs(X, Y, m, n, dp): # base case if (m == 0 or n == 0): return 0 # if the same state has already been # computed if (dp[m - 1][n - 1] != -1): return dp[m - 1][n - 1] # if equal, then we store the value of the # function call if (X[m - 1] == Y[n - 1]): # store it in arr to avoid further repetitive # work in future function calls dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp) return dp[m - 1][n - 1] else : # store it in arr to avoid further repetitive # work in future function calls dp[m - 1][n - 1] = max(lcs(X, Y, m, n - 1, dp), lcs(X, Y, m - 1, n, dp)) return dp[m - 1][n - 1]# Driver CodeX = "AGGTAB"Y = "GXTXAYB"m = len(X)n = len(Y)dp = [[-1 for i in range(maximum)] for i in range(m)]print("Length of LCS:", lcs(X, Y, m, n, dp))# This code is contributed by Mohit Kumar |
C#
// C# program to memoize// recursive implementation of LCS problem using System;class GFG {static readonly int maximum = 1000;// Returns length of LCS for // X[0..m-1], Y[0..n-1] // memoization applied in // recursive solution static int lcs(String X, String Y, int m, int n, int [,]dp){ // base case if (m == 0 || n == 0) { return 0; } // if the same state has already been // computed if (dp[m - 1, n - 1] != -1) { return dp[m - 1, n - 1]; } // if equal, then we store the value // of the function call if (X[m - 1] == Y[n - 1]) { // store it in arr to avoid // further repetitive work // in future function calls dp[m - 1, n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp); return dp[m - 1, n - 1]; } else { // store it in arr to avoid // further repetitive work // in future function calls 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]; }}// Driver Code 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]; // assign -1 to all positions for(int i = 0; i < m; i++) { for(int j = 0; j < maximum; j++) { dp[i, j] = -1; } } Console.WriteLine("Length of LCS: " + lcs(X, Y, m, n, dp));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to memoize// recursive implementation of LCS problemlet maximum = 1000;// Returns length of LCS for X[0..m-1], Y[0..n-1] */// memoization applied in recursive solutionfunction lcs(X,Y,m,n,dp){ // base case if (m == 0 || n == 0) { return 0; } // if the same state has already been // computed if (dp[m - 1][n - 1] != -1) { return dp[m - 1][n - 1]; } // if equal, then we store the value of the // function call if (X[m-1] == Y[n-1]) { // store it in arr to avoid further repetitive // work in future function calls dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp); return dp[m - 1][n - 1]; } else { // store it in arr to avoid further repetitive // work in future function calls 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]; }}// Driver Codelet X = "AGGTAB";let Y = "GXTXAYB";let m = X.length;let n = Y.length;let dp=new Array(m);for(let i=0;i<dp.length;i++){ dp[i]=new Array(maximum); for(let j=0;j<dp[i].length;j++) { dp[i][j]=-1; }}document.write("Length of LCS: " + lcs(X, Y, m, n, dp)); // This code is contributed by avanitrachhadiya2155</script> |
Output
Length of LCS: 4
complexity Analysis:
- Time Complexity: O(N * M), where N and M are lengths of the first and second string respectively.
- Auxiliary Space: (N * M)
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