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Longest Common Substring (Space optimized DP solution)
• Difficulty Level : Hard
• Last Updated : 26 Apr, 2021

Given two strings ‘X’ and ‘Y’, find the length of longest common substring. Expected space complexity is linear.
Examples :

```Input : X = "GeeksforGeeks", Y = "GeeksQuiz"
Output : 5
The longest common substring is "Geeks" and is of
length 5.

Input : X = "abcdxyz", Y = "xyzabcd"
Output : 4
The longest common substring is "abcd" and is of
length 4.``` We have discussed Dynamic programming based solution for Longest common substring. The auxiliary space used by the solution is O(m*n), where m and n are lengths of string X and Y. The space used by solution can be reduced to O(2*n).
Suppose we are at position mat[i][j]. Now if X[i-1] == Y[j-1], then we add the value of mat[i-1][j-1] to our result. That is we add value from previous row and value for all other rows below the previous row are never used. So, at a time we are using only two consecutive rows. This observation can be used to reduce the space required to find length of longest common substring.
Instead of creating a matrix of size m*n, we create a matrix of size 2*n. A variable currRow is used to represent that either row 0 or row 1 of this matrix is currently used to find length. Initially row 0 is used as current row for the case when length of string X is zero. At the end of each iteration, current row is made previous row and previous row is made new current row.

## C++

 `// Space optimized CPP implementation of longest``// common substring.``#include ``using` `namespace` `std;` `// Function to find longest common substring.``int` `LCSubStr(string X, string Y)``{``    ``// Find length of both the strings.``    ``int` `m = X.length();``    ``int` `n = Y.length();` `    ``// Variable to store length of longest``    ``// common substring.``    ``int` `result = 0;` `    ``// Matrix to store result of two``    ``// consecutive rows at a time.``    ``int` `len[n];` `    ``// Variable to represent which row of``    ``// matrix is current row.``    ``int` `currRow = 0;` `    ``// For a particular value of i and j,``    ``// len[currRow][j] stores length of longest``    ``// common substring in string X[0..i] and Y[0..j].``    ``for` `(``int` `i = 0; i <= m; i++) {``        ``for` `(``int` `j = 0; j <= n; j++) {``            ``if` `(i == 0 || j == 0) {``                ``len[currRow][j] = 0;``            ``}``            ``else` `if` `(X[i - 1] == Y[j - 1]) {``                ``len[currRow][j] = len[1 - currRow][j - 1] + 1;``                ``result = max(result, len[currRow][j]);``            ``}``            ``else` `{``                ``len[currRow][j] = 0;``            ``}``        ``}` `        ``// Make current row as previous row and previous``        ``// row as new current row.``        ``currRow = 1 - currRow;``    ``}` `    ``return` `result;``}` `int` `main()``{``    ``string X = ``"GeeksforGeeks"``;``    ``string Y = ``"GeeksQuiz"``;` `    ``cout << LCSubStr(X, Y);``    ``return` `0;``}`

## Java

 `// Space optimized CPP implementation of``// longest common substring.``import` `java.io.*;``import` `java.util.*;` `public` `class` `GFG {``    ` `    ``// Function to find longest``    ``// common substring.``    ``static` `int` `LCSubStr(String X, String Y)``    ``{``        ` `        ``// Find length of both the strings.``        ``int` `m = X.length();``        ``int` `n = Y.length();``    ` `        ``// Variable to store length of longest``        ``// common substring.``        ``int` `result = ``0``;``    ` `        ``// Matrix to store result of two``        ``// consecutive rows at a time.``        ``int` `[][]len = ``new` `int``[``2``][n];``    ` `        ``// Variable to represent which row of``        ``// matrix is current row.``        ``int` `currRow = ``0``;``    ` `        ``// For a particular value of``        ``// i and j, len[currRow][j]``        ``// stores length of longest``        ``// common substring in string``        ``// X[0..i] and Y[0..j].``        ``for` `(``int` `i = ``0``; i < m; i++) {``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``if` `(i == ``0` `|| j == ``0``) {``                    ``len[currRow][j] = ``0``;``                ``}``                ``else` `if` `(X.charAt(i - ``1``) ==``                              ``Y.charAt(j - ``1``))``                ``{``                    ``len[currRow][j] =``                      ``len[(``1` `- currRow)][(j - ``1``)]``                                             ``+ ``1``;``                    ``result = Math.max(result,``                                ``len[currRow][j]);``                ``}``                ``else``                ``{``                    ``len[currRow][j] = ``0``;``                ``}``            ``}``    ` `            ``// Make current row as previous``            ``// row and previous row as``            ``// new current row.``            ``currRow = ``1` `- currRow;``        ``}``    ` `        ``return` `result;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``String X = ``"GeeksforGeeks"``;``        ``String Y = ``"GeeksQuiz"``;``    ` `        ``System.out.print(LCSubStr(X, Y));``    ``}``}` `// This code is contributed by``// Manish Shaw (manishshaw1)`

## Python3

 `# Space optimized Python3 implementation ``# of longest common substring.``import` `numpy as np` `# Function to find longest common substring.``def` `LCSubStr(X, Y) :` `    ``# Find length of both the strings.``    ``m ``=` `len``(X)``    ``n ``=` `len``(Y)` `    ``# Variable to store length of``    ``# longest common substring.``    ``result ``=` `0` `    ``# Matrix to store result of two``    ``# consecutive rows at a time.``    ``len_mat ``=` `np.zeros((``2``, n))` `    ``# Variable to represent which row ``    ``# of matrix is current row.``    ``currRow ``=` `0` `    ``# For a particular value of i and j,``    ``# len_mat[currRow][j] stores length of``    ``# longest common substring in string``    ``# X[0..i] and Y[0..j].``    ``for` `i ``in` `range``(m) :``        ``for` `j ``in` `range``(n) :``                        ` `            ``if` `(i ``=``=` `0` `| j ``=``=` `0``) :``                ``len_mat[currRow][j] ``=` `0``            ` `            ``elif` `(X[i ``-` `1``] ``=``=` `Y[j ``-` `1``]) :``                                ` `                ``len_mat[currRow][j] ``=` `len_mat[``1` `-` `currRow][j ``-` `1``] ``+` `1``                ``result ``=` `max``(result, len_mat[currRow][j])``            ` `            ``else` `:``                ``len_mat[currRow][j] ``=` `0``            ` `        ``# Make current row as previous row and``        ``# previous row as new current row.``        ``currRow ``=` `1` `-` `currRow` `    ``return` `result` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``X ``=` `"GeeksforGeeks"``    ``Y ``=` `"GeeksQuiz"` `    ``print``(LCSubStr(X, Y))` `# This code is contributed by Ryuga`

## C#

 `// Space optimized C# implementation``// of longest common substring.``using` `System;``using` `System.Collections.Generic;``class` `GFG {``    ` `    ``// Function to find longest``    ``// common substring.``    ``static` `int` `LCSubStr(``string` `X, ``string` `Y)``    ``{``        ` `        ``// Find length of both the strings.``        ``int` `m = X.Length;``        ``int` `n = Y.Length;``    ` `        ``// Variable to store length of longest``        ``// common substring.``        ``int` `result = 0;``    ` `        ``// Matrix to store result of two``        ``// consecutive rows at a time.``        ``int` `[,]len = ``new` `int``[2,n];``    ` `        ``// Variable to represent which row of``        ``// matrix is current row.``        ``int` `currRow = 0;``    ` `        ``// For a particular value of``        ``// i and j, len[currRow][j]``        ``// stores length of longest``        ``// common substring in string``        ``// X[0..i] and Y[0..j].``        ``for` `(``int` `i = 0; i < m; i++) {``            ``for` `(``int` `j = 0; j < n; j++) {``                ``if` `(i == 0 || j == 0) {``                    ``len[currRow,j] = 0;``                ``}``                ``else` `if` `(X[i - 1] == Y[j - 1]) {``                    ``len[currRow,j] = len[(1 - currRow),``                                          ``(j - 1)] + 1;``                    ``result = Math.Max(result, len[currRow, j]);``                ``}``                ``else``                ``{``                    ``len[currRow,j] = 0;``                ``}``            ``}``    ` `            ``// Make current row as previous``            ``// row and previous row as``            ``// new current row.``            ``currRow = 1 - currRow;``        ``}``    ` `        ``return` `result;``    ``}``    ` `    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``string` `X = ``"GeeksforGeeks"``;``        ``string` `Y = ``"GeeksQuiz"``;``    ` `        ``Console.Write(LCSubStr(X, Y));``    ``}``}` `// This code is contributed by``// Manish Shaw (manishshaw1)`

## PHP

 ``

## Javascript

 ``
Output :
`5`

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

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