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# A Space Optimized Solution of LCS

• Difficulty Level : Medium
• Last Updated : 09 Sep, 2021

Given two strings, find the length of the 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 a typical dynamic programming-based solution for LCS. We can optimize the space used by LCS problem. We know the recurrence relationship of the LCS problem is

## CPP

 `/* 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];``}`

## Java

 `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``[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];``    ``}``}` `// This code is contributed by rajsanghavi9.`

How to find the length of LCS is O(n) auxiliary space?

We strongly recommend that you click here and practice it, before moving on to the solution.
One important observation in the above simple implementation is, in each iteration of the outer loop we only need values from all columns of the previous row. So there is no need to store all rows in our DP matrix, we can just store two rows at a time and use them. In that way, used space will be reduced from L[m+1][n+1] to L[2][n+1]. Below is the implementation of the above idea.

## C++

 `// Space optimized C++ implementation``// of LCS problem``#include``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-1] == 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 code``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 Code``    ``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.`

## Python3

 `# Space optimized Python``# implementation of LCS problem` `# Returns length of LCS for``# X[0..m-1], Y[0..n-1]``def` `lcs(X, Y):``    ` `    ``# Find lengths of two strings``    ``m ``=` `len``(X)``    ``n ``=` `len``(Y)` `    ``L ``=` `[[``0` `for` `i ``in` `range``(n``+``1``)] ``for` `j ``in` `range``(``2``)]` `    ``# Binary index, used to index current row and``    ``# previous row.``    ``bi ``=` `bool``    ` `    ``for` `i ``in` `range``(m):``        ``# Compute current binary index``        ``bi ``=` `i&``1` `        ``for` `j ``in` `range``(n``+``1``):``            ``if` `(i ``=``=` `0` `or` `j ``=``=` `0``):``                ``L[bi][j] ``=` `0` `            ``elif` `(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 Code``X ``=` `"AGGTAB"``Y ``=` `"GXTXAYB"` `print``(``"Length of LCS is"``, lcs(X, Y))` `# This code is contributed by Soumen Ghosh.`

## C#

 `// C# Code for A Space``// Optimized Solution of LCS``using` `System;` `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[i - 1] == Y[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 Code``    ``public` `static` `void` `Main()``    ``{``        ``string` `X = ``"AGGTAB"``;``        ``string` `Y = ``"GXTXAYB"``;``    ` `        ``Console.Write(``"Length of LCS is "` `+``                                ``lcs(X, Y));``    ``}``}` `// This code is contributed``// by shiv_bhakt.`

## PHP

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## Javascript

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Output:

`Length of LCS is 4`

Time Complexity : O(m*n)
Auxiliary Space : O(n)