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.

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 * m^{2})

**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 LCSuff(X, Y, m, n) = LCSuff(X, Y, m-1, n-1) + 1 if X[m-1] = Y[n-1] 0 Otherwise (if X[m-1] != Y[n-1]) 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 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; // A utility function to find maximum of two integers int max(int a, int b) { return (a > b)? a : b; } /* 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. 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 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++) { 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

Output:

Length of Longest Common Substring is 10

Time Complexity: O(m*n)

Auxiliary Space: O(m*n)

**References:** http://en.wikipedia.org/wiki/Longest_common_substring_problem

The longest substring can also be solved in O(n+m) time using Suffix Tree. We will be covering Suffix Tree based solution in a separate post.

**Exercise: **The above solution prints only length of the longest common substring. Extend the solution to print the substring also.

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