Given two strings, find the length of longest subsequence present in both of them.

**Examples:**

LCS for input Sequences “**ABCDGH**” and “**AEDFHR**” is “**ADH**” of length **3**.

LCS for input Sequences “**AGGTAB**” and “**GXTXAYB**” is “**GTAB**” of length **4**.

We have discussed typical dynamic programming based solution for LCS. We can optimize space used by lcs problem. We know recurrence relation of LCS problem is

/* Returns length of LCS for X[0..m-1], Y[0..n-1] */ int lcs(string &X, string &Y) { int m = X.length(), n = Y.length(); int L[m+1][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] */ 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[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]); } } /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */ return L[m][n]; }

**How to find length of LCS in O(n) auxiliary space?**

We strongly recommend that you click here and practice it, before moving on to the solution.

One important observation in above simple implementation is, in each iteration of outer loop we only, need **values from all columns of previous row**. So there is no need of storing all rows in our DP matrix, we can just store two rows at a time and use them, in that way used space will reduce from L[m+1][n+1] to L[2][n+1]. Below is C++ implementation of this idea.

## C++

/* Space optimized C++ implementation of LCS problem */ #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) { // Find lengths of two strings int m = X.length(), n = Y.length(); int L[2][n+1]; // Binary index, used to index current row and // previous row. bool bi; for (int i=0; i<=m; i++) { // Compute current binary index bi = i&1; for (int j=0; j<=n; j++) { if (i == 0 || j == 0) L[bi][j] = 0; else if (X[i] == Y[j-1]) L[bi][j] = L[1-bi][j-1] + 1; else L[bi][j] = max(L[1-bi][j], L[bi][j-1]); } } /* Last filled entry contains length of LCS for X[0..n-1] and Y[0..m-1] */ return L[bi][n]; } /* Driver program to test above function */ int main() { string X = "AGGTAB"; string Y = "GXTXAYB"; printf("Length of LCS is %d\n", lcs(X, Y)); return 0; }

## Java

// JAVA Code for A Space Optimized // Solution of LCS class GFG { /* Returns length of LCS for X[0..m-1], Y[0..n-1] */ public static int lcs(String X, String Y) { // Find lengths of two strings int m = X.length(), n = Y.length(); int L[][] = new int[2][n+1]; // Binary index, used to index current // row and previous row. int bi=0; for (int i = 0; i <= m; i++) { // Compute current binary index bi = i & 1; for (int j = 0; j <= n; j++) { if (i == 0 || j == 0) L[bi][j] = 0; else if (X.charAt(i-1) == Y.charAt(j-1)) L[bi][j] = L[1-bi][j-1] + 1; else L[bi][j] = Math.max(L[1-bi][j], L[bi][j-1]); } } /* Last filled entry contains length of LCS for X[0..n-1] and Y[0..m-1] */ return L[bi][n]; } /* Driver program to test above function */ public static void main(String[] args) { String X = "AGGTAB"; String Y = "GXTXAYB"; System.out.println("Length of LCS is " + lcs(X, Y)); } } // This code is contributed by Arnav Kr. Mandal.

Output:

Length of LCS is 4

Time Complexity : O(m*n)

Auxiliary Space : O(n)

This article is contributed **Shivam Mittal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above