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.

longest-common-substring

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++

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// 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;
}

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Java














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// 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) 



















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Python3














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# 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 



















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C#














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// 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) 



















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PHP














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<?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) 


?> 



















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Output :


5





Time Complexity:  O(m*n)

Auxiliary Space:  O(n)


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