Longest Common Substring (Space optimized DP solution)

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 <bits/stdc++.h>
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[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 = 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

<?php
// Space optimized PHP implementation
// of longest common substring.

// Function to find
// longest common substring.
function LCSubStr($X, $Y)
{
    // Find length of
    // both the strings.
    $m = strlen($X);
    $n = strlen($Y);

    // Variable to store length
    // of longest common substring.
    $result = 0;

    // Matrix to store result of two
    // consecutive rows at a time.
    $len = array(array(), array(), );

    // Variable to represent which
    // row of matrix is current row.
    $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 ($i = 0; $i <= $m; $i++)
    {
        for ($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;
}

// Driver Code
$X = "GeeksforGeeks";
$Y = "GeeksQuiz";

print (LCSubStr($X, $Y));

// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Javascript

<script>

// Space optimized CPP implementation of longest
// common substring.

// Function to find longest common substring.
function LCSubStr(X, Y)
{
    // Find length of both the strings.
    var m = X.length;
    var n = Y.length;

    // Variable to store length of longest
    // common substring.
    var result = 0;

    // Matrix to store result of two
    // consecutive rows at a time.
    var len = Array.from(Array(2), ()=> Array(n));

    // Variable to represent which row of
    // matrix is current row.
    var 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 (var i = 0; i <= m; i++) {
        for (var 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
var X = "GeeksforGeeks";
var Y = "GeeksQuiz";
document.write( LCSubStr(X, Y));

// This code is contributed by itsok.
</script>

Output :

5

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