Longest Common Substring | DP-29

Given two strings ‘X’ and ‘Y’, find the length of the longest common substring.

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.

Input : X = “zxabcdezy”, y = “yzabcdezx”
Output : 6
The longest common substring is “abcdez” and is of length 6.

longest-common-substring

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Let m and n be the lengths of first and second strings respectively.

A simple solution is to one by one consider all substrings of first string and for every substring check if it is a substring in second string. Keep track of the maximum length substring. There will be O(m^2) substrings and we can find whether a string is subsring on another string in O(n) time (See this). So overall time complexity of this method would be O(n * m²)

Dynamic Programming can be used to find the longest common substring in O(m*n) time. The idea is to find length of the longest common suffix for all substrings of both strings and store these lengths in a table.

The longest common suffix has following optimal substructure property.

If last characters match, then we reduce both lengths by 1
LCSuff(X, Y, m, n) = LCSuff(X, Y, m-1, n-1) + 1 if X[m-1] = Y[n-1]
If last characters do not match, then result is 0, i.e.,
LCSuff(X, Y, m, n) = 0 if (X[m-1] != Y[n-1])

Now we consider suffixes of different substrings ending at different indexes.
The maximum length Longest Common Suffix is the longest common substring.
LCSubStr(X, Y, m, n) = Max(LCSuff(X, Y, i, j)) where 1 <= i <= m and 1 <= j <= n

Following is the iterative implementation of the above solution.

C++

/* Dynamic Programming solution to find length of the  
   longest common substring */
#include<iostream> 
#include<string.h> 
using namespace std; 
  
/* Returns length of longest common substring of X[0..m-1]  
   and Y[0..n-1] */
int LCSubStr(char *X, char *Y, int m, int n) 
{ 
    // Create a table to store lengths of longest 
    // common suffixes of substrings.   Note that 
    // LCSuff[i][j] contains length of longest 
    // common suffix of X[0..i-1] and Y[0..j-1].  
  
    int LCSuff[m+1][n+1]; 
    int result = 0;  // To store length of the  
                     // longest common substring 
  
    /* Following steps build LCSuff[m+1][n+1] in 
        bottom up fashion. */
    for (int i=0; i<=m; i++) 
    { 
        for (int j=0; j<=n; j++) 
        { 
  
            // The first row and first column  
            // entries have no logical meaning,  
            // they are used only for simplicity  
            // of program 
            if (i == 0 || j == 0) 
                LCSuff[i][j] = 0; 
  
            else if (X[i-1] == Y[j-1]) 
            { 
                LCSuff[i][j] = LCSuff[i-1][j-1] + 1; 
                result = max(result, LCSuff[i][j]); 
            } 
            else LCSuff[i][j] = 0; 
        } 
    } 
    return result; 
} 
  
/* Driver program to test above function */
int main() 
{ 
    char X[] = "OldSite:GeeksforGeeks.org"; 
    char Y[] = "NewSite:GeeksQuiz.com"; 
  
    int m = strlen(X); 
    int n = strlen(Y); 
  
    cout << "Length of Longest Common Substring is " 
         << LCSubStr(X, Y, m, n); 
    return 0; 
} 

Java

//  Java implementation of finding length of longest  
// Common substring using Dynamic Programming 
public class LongestCommonSubSequence  
{ 
    /*  
       Returns length of longest common substring   
       of X[0..m-1] and Y[0..n-1]  
    */
    static int LCSubStr(char X[], char Y[], int m, int n)  
    { 
        // Create a table to store lengths of longest common suffixes of 
        // substrings. Note that LCSuff[i][j] contains length of longest 
        // common suffix of X[0..i-1] and Y[0..j-1]. The first row and 
        // first column entries have no logical meaning, they are used only 
        // for simplicity of program 
        int LCStuff[][] = new int[m + 1][n + 1]; 
        int result = 0;  // To store length of the longest common substring 
          
        // Following steps build LCSuff[m+1][n+1] in bottom up fashion 
        for (int i = 0; i <= m; i++)  
        { 
            for (int j = 0; j <= n; j++)  
            { 
                if (i == 0 || j == 0) 
                    LCStuff[i][j] = 0; 
                else if (X[i - 1] == Y[j - 1]) 
                { 
                    LCStuff[i][j] = LCStuff[i - 1][j - 1] + 1; 
                    result = Integer.max(result, LCStuff[i][j]); 
                }  
                else
                    LCStuff[i][j] = 0; 
            } 
        } 
        return result; 
    } 
      
    // Driver Program to test above function 
    public static void main(String[] args)  
    { 
        String X = "OldSite:GeeksforGeeks.org"; 
        String Y = "NewSite:GeeksQuiz.com"; 
  
        int m = X.length(); 
        int n = Y.length(); 
  
        System.out.println("Length of Longest Common Substring is "
                + LCSubStr(X.toCharArray(), Y.toCharArray(), m, n)); 
    } 
} 
  
// This code is contributed by Sumit Ghosh 

Python3

# Python3 implementation of Finding  
# Length of Longest Common Substring  
  
# Returns length of longest common  
# substring of X[0..m-1] and Y[0..n-1]  
def LCSubStr(X, Y, m, n): 
      
    # Create a table to store lengths of 
    # longest common suffixes of substrings.  
    # Note that LCSuff[i][j] contains the  
    # length of longest common suffix of  
    # X[0...i-1] and Y[0...j-1]. The first 
    # row and first column entries have no 
    # logical meaning, they are used only 
    # for simplicity of the program. 
      
    # LCSuff is the table with zero  
    # value initially in each cell 
    LCSuff = [[0 for k in range(n+1)] for l in range(m+1)] 
      
    # To store the length of  
    # longest common substring 
    result = 0 
  
    # Following steps to build 
    # LCSuff[m+1][n+1] in bottom up fashion 
    for i in range(m + 1): 
        for j in range(n + 1): 
            if (i == 0 or j == 0): 
                LCSuff[i][j] = 0
            elif (X[i-1] == Y[j-1]): 
                LCSuff[i][j] = LCSuff[i-1][j-1] + 1
                result = max(result, LCSuff[i][j]) 
            else: 
                LCSuff[i][j] = 0
    return result 
  
# Driver Program to test above function 
X = 'OldSite:GeeksforGeeks.org'
Y = 'NewSite:GeeksQuiz.com'
  
m = len(X) 
n = len(Y) 
  
print('Length of Longest Common Substring is', 
                      LCSubStr(X, Y, m, n)) 
  
# This code is contributed by Soumen Ghosh 

C#

// C# implementation of finding length of longest 
// Common substring using Dynamic Programming 
using System; 
  
class GFG { 
       
    // Returns length of longest common  
    // substring of X[0..m-1] and Y[0..n-1]  
    static int LCSubStr(string X, string Y, 
                                 int m, int n) 
    { 
          
        // Create a table to store lengths of  
        // longest common suffixes of substrings. 
        // Note that LCSuff[i][j] contains length 
        // of longest common suffix of X[0..i-1]  
        // and Y[0..j-1]. The first row and first 
        // column entries have no logical meaning, 
        // they are used only for simplicity of  
        // program 
        int[, ] LCStuff = new int[m + 1, n + 1]; 
          
        // To store length of the longest common 
        // substring 
        int result = 0;  
  
        // Following steps build LCSuff[m+1][n+1]  
        // in bottom up fashion 
        for (int i = 0; i <= m; i++) { 
            for (int j = 0; j <= n; j++) { 
                if (i == 0 || j == 0) 
                    LCStuff[i, j] = 0; 
                else if (X[i - 1] == Y[j - 1]) { 
                    LCStuff[i, j] =  
                            LCStuff[i - 1, j - 1] + 1; 
                              
                    result = Math.Max(result,  
                                      LCStuff[i, j]); 
                } 
                else
                    LCStuff[i, j] = 0; 
            } 
        } 
          
        return result; 
    } 
  
    // Driver Program to test above function 
    public static void Main() 
    { 
        String X = "OldSite:GeeksforGeeks.org"; 
        String Y = "NewSite:GeeksQuiz.com"; 
  
        int m = X.Length; 
        int n = Y.Length; 
  
        Console.Write("Length of Longest Common"
        + " Substring is " + LCSubStr(X, Y, m, n)); 
    } 
  
} 
  
// This code is contributed by Sam007. 

PHP

<?php  
// Dynamic Programming solution to find  
// length of the longest common substring  
  
// Returns length of longest common  
// substring of X[0..m-1] and Y[0..n-1]  
function LCSubStr($X, $Y, $m, $n) 
{ 
    // Create a table to store lengths of  
    // longest common suffixes of substrings.  
    // Notethat LCSuff[i][j] contains length  
    // of longest common suffix of X[0..i-1]  
    // and Y[0..j-1]. The first row and 
    // first column entries have no logical  
    // meaning, they are used only for  
    // simplicity of program 
    $LCSuff = array_fill(0, $m + 1, 
              array_fill(0, $n + 1, NULL)); 
    $result = 0; // To store length of the  
                 // longest common substring 
  
    // Following steps build LCSuff[m+1][n+1]  
    // in bottom up fashion.  
    for ($i = 0; $i <= $m; $i++) 
    { 
        for ($j = 0; $j <= $n; $j++) 
        { 
            if ($i == 0 || $j == 0) 
                $LCSuff[$i][$j] = 0; 
  
            else if ($X[$i - 1] == $Y[$j - 1]) 
            { 
                $LCSuff[$i][$j] = $LCSuff[$i - 1][$j - 1] + 1; 
                $result = max($result,  
                              $LCSuff[$i][$j]); 
            } 
            else $LCSuff[$i][$j] = 0; 
        } 
    } 
    return $result; 
} 
  
// Driver Code  
$X = "OldSite:GeeksforGeeks.org"; 
$Y = "NewSite:GeeksQuiz.com"; 
  
$m = strlen($X); 
$n = strlen($Y); 
  
echo "Length of Longest Common Substring is " .  
                      LCSubStr($X, $Y, $m, $n); 
                        
// This code is contributed by ita_c 
?> 

Output:

Length of Longest Common Substring is 10

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

Another approach: (Using recursion)
Here is the recursive solution of above approach.

C++

// C++ program using to find length of the  
// longest common substring  recursion 
#include <iostream> 
  
using namespace std; 
  
string X,Y; 
  
// Returns length of function for longest common   
// substring of X[0..m-1] and Y[0..n-1]  
int lcs(int i, int j, int count)  
{ 
      
    if (i == 0 || j == 0) 
        return count; 
          
    if (X[i-1] == Y[j-1]) { 
        count = lcs(i - 1, j - 1, count + 1); 
    } 
        count = max(count, max(lcs( i, j - 1, 0), lcs( i - 1, j, 0))); 
    return count; 
} 
  
// Driver code 
int main() 
{ 
int n,m; 
  
X = "abcdxyz";  
Y = "xyzabcd"; 
  
n=X.size(); 
m=Y.size(); 
  
cout<<lcs(n,m,0); 
  
    return 0; 
} 

Java

// Java program using to find length of the  
// longest common substring recursion  
  
class GFG { 
  
    static String X, Y; 
// Returns length of function for longest common  
// substring of X[0..m-1] and Y[0..n-1]  
    static int lcs(int i, int j, int count) { 
  
        if (i == 0 || j == 0) { 
            return count; 
        } 
  
        if (X.charAt(i - 1) == Y.charAt(j - 1)) { 
            count = lcs(i - 1, j - 1, count + 1); 
        } 
        count = Math.max(count, Math.max(lcs(i, j - 1, 0), 
                            lcs(i - 1, j, 0))); 
        return count; 
    } 
// Driver code  
    public static void main(String[] args) { 
        int n, m; 
        X = "abcdxyz"; 
        Y = "xyzabcd"; 
  
        n = X.length(); 
        m = Y.length(); 
  
        System.out.println(lcs(n, m, 0)); 
    } 
} 
// This code is contributed by Rajput-JI  

Python3

# Python3 program using to find length of  
# the longest common substring recursion  
  
# Returns length of function for longest  
# common substring of X[0..m-1] and Y[0..n-1]  
def lcs(i, j, count) :  
      
    if (i == 0 or j == 0) :  
        return count  
          
    if (X[i - 1] == Y[j - 1]) : 
        count = lcs(i - 1, j - 1, count + 1)  
      
    count = max(count, max(lcs( i, j - 1, 0),  
                           lcs( i - 1, j, 0)))  
  
    return count 
  
# Driver code  
if __name__ == "__main__" : 
  
    X = "abcdxyz"
    Y = "xyzabcd"
  
    n = len(X) 
    m = len(Y) 
  
    print(lcs(n, m, 0))  
  
# This code is contributed by Ryuga 

C#

// C# program using to find length  
// of the longest common substring 
// recursion  
using System; 
  
class GFG 
{  
static String X, Y; 
  
// Returns length of function for  
// longest common substring of 
// X[0..m-1] and Y[0..n-1]  
static int lcs(int i, int j, int count)  
{ 
  
    if (i == 0 || j == 0) 
    { 
        return count; 
    } 
  
    if (X[i - 1] == Y[j - 1])  
    { 
        count = lcs(i - 1, j - 1, count + 1); 
    } 
    count = Math.Max(count, Math.Max(lcs(i, j - 1, 0),  
                                     lcs(i - 1, j, 0))); 
    return count; 
} 
  
// Driver code  
public static void Main()  
{ 
    int n, m; 
    X = "abcdxyz"; 
    Y = "xyzabcd"; 
  
    n = X.Length; 
    m = Y.Length; 
  
    Console.Write(lcs(n, m, 0)); 
} 
} 
  
// This code is contributed by Rajput-JI  

PHP

<?php 
// PHP program using to find length of the  
// longest common substring recursion  
  
// Returns length of function for  
// longest common substring of 
// X[0..m-1] and Y[0..n-1]  
function lcs($i, $j, $count, &$X, &$Y)  
{  
    if ($i == 0 || $j == 0)  
        return $count;  
          
    if ($X[$i - 1] == $Y[$j - 1])  
    {  
        $count = lcs($i - 1, $j - 1,  
                     $count + 1, $X, $Y);  
    }  
        $count = max($count, lcs($i, $j - 1, 0, $X, $Y),  
                             lcs($i - 1, $j, 0, $X, $Y));  
    return $count;  
}  
  
// Driver code  
$X = "abcdxyz";  
$Y = "xyzabcd";  
  
$n = strlen($X);  
$m = strlen($Y);  
  
echo lcs($n, $m, 0, $X, $Y);  
  
// This code is contributed 
// by rathbhupendra 
?> 