# Longest Common Substring | DP-29

• Difficulty Level : Medium
• Last Updated : 21 Jun, 2022

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

Examples :

Input : X = “GeeksforGeeks”, y = “GeeksQuiz”
Output : 5
Explanation:
The longest common substring is “Geeks” and is of length 5.

Input : X = “abcdxyz”, y = “xyzabcd”
Output :
Explanation:
The longest common substring is “abcd” and is of length 4.

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

Recommended Practice

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

A simple solution is to one by one consider all substrings of the first string and for every substring check if it is a substring in the second string. Keep track of the maximum length substring. There will be O(m^2) substrings and we can find whether a string is substring 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 the 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 #include 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 code 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 class GFG {     /*        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];                 // 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 = Integer.max(result,                                          LCStuff[i][j]);                 }                 else                     LCStuff[i][j] = 0;             }         }         return result;     }       // Driver Code     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(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 Code 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 Code     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.



## Javascript



Output

Length of Longest Common Substring is 10

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

Another approach: (Space optimized approach).
In the above approach, we are only using the last row of the 2-D array only, hence we can optimize the space by using
a 2-D array of dimension 2*(min(n,m)).

Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach #include using namespace std;   // Function to find the length of the // longest LCS int LCSubStr(string s, string t, int n, int m) {         // Create DP table     int dp[2][m + 1];     int res = 0;       for (int i = 1; i <= n; i++) {         for (int j = 1; j <= m; j++) {             if (s[i - 1] == t[j - 1]) {                 dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;                 if (dp[i % 2][j] > res)                     res = dp[i % 2][j];             }             else                 dp[i % 2][j] = 0;         }     }     return res; }   // Driver Code int main() {     string X = "OldSite:GeeksforGeeks.org";     string Y = "NewSite:GeeksQuiz.com";       int m = X.length();     int n = Y.length();       cout << LCSubStr(X, Y, m, n);     return 0;     cout << "GFG!";     return 0; }   // This code is contributed by rajsanghavi9.

## Java

 // Java implementation of the above approach   class GFG {         // Function to find the length of the     // longest LCS     static int LCSubStr(String s,String t,                         int n,int m)     {                  // Create DP table         int dp[][]=new int[2][m+1];         int res=0;                for(int i=1;i<=n;i++)         {             for(int j=1;j<=m;j++)             {                 if(s.charAt(i-1)==t.charAt(j-1))                 {                     dp[i%2][j]=dp[(i-1)%2][j-1]+1;                     if(dp[i%2][j]>res)                         res=dp[i%2][j];                 }                 else dp[i%2][j]=0;             }         }         return res;     }         // Driver Code     public static void main (String[] args)     {         String X="OldSite:GeeksforGeeks.org";         String Y="NewSite:GeeksQuiz.com";                   int m=X.length();         int n=Y.length();                   // Function call         System.out.println(LCSubStr(X,Y,m,n));               } }

## Python3

 # Python implementation of the above approach   # Function to find the length of the # longest LCS def LCSubStr(s, t, n, m):         # Create DP table     dp = [[0 for i in range(m + 1)] for j in range(2)]     res = 0           for i in range(1,n + 1):         for j in range(1,m + 1):             if(s[i - 1] == t[j - 1]):                 dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1                 if(dp[i % 2][j] > res):                     res = dp[i % 2][j]             else:                 dp[i % 2][j] = 0     return res   # Driver Code X = "OldSite:GeeksforGeeks.org" Y = "NewSite:GeeksQuiz.com" m = len(X) n = len(Y)   # Function call print(LCSubStr(X,Y,m,n))   # This code is contributed by avanitrachhadiya2155

## C#

 // C# implementation of the above approach using System; public class GFG {     // Function to find the length of the   // longest LCS   static int LCSubStr(string s,string t,                       int n,int m)   {        // Create DP table     int[,] dp = new int[2, m + 1];     int res = 0;       for(int i = 1; i <= n; i++)     {       for(int j = 1; j <= m; j++)       {         if(s[i - 1] == t[j - 1])         {           dp[i % 2, j] = dp[(i - 1) % 2, j - 1] + 1;           if(dp[i % 2, j] > res)             res = dp[i % 2, j];         }         else dp[i % 2, j] = 0;       }     }     return res;   }     // Driver Code   static public void Main (){     string X = "OldSite:GeeksforGeeks.org";     string Y = "NewSite:GeeksQuiz.com";       int m = X.Length;     int n = Y.Length;       // Function call     Console.WriteLine(LCSubStr(X,Y,m,n));   } }   // This code is contributed by rag2127

## Javascript



Output

10

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

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

## C++

 // C++ program using to find length of the // longest common substring  recursion #include   using namespace std;   string X, Y;   // Returns length of function f // or 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



## Javascript



Output

4

#### Maximum Space Optimization:

1. In this method, we will use recursion to find the longest prefix of all the possible substrings.
2. Let  gives the length of the longest common suffix starting from indices i, j of strings X, Y respectively.
3. Then the function can be defined as :
4. In this recursion we can see that the function has only one dependency, so that means we can get away with memorizing just the previous computation if we do our computation in a specific order.
5. Consider the following table where we memorize the solutions:

We need to find the solution diagonally upwards. In this particular example:

• first diagonal
• (4, 0)
• second diagonal
• (4, 1)
• (3, 0)
• third diagonal
• (4, 2)
• (3, 1)
• (2, 0)

Like this, we need to remember only the previous computation.

## Python3

 # Python code for the above approach from functools import lru_cache from operator import itemgetter   def longest_common_substring(x: str, y: str) -> (int, int, int):           # function to find the longest common substring       # Memorizing with maximum size of the memory as 1     @lru_cache(maxsize=1)            # function to find the longest common prefix     def longest_common_prefix(i: int, j: int) -> int:                 if 0 <= i < len(x) and 0 <= j < len(y) and x[i] == y[j]:             return 1 + longest_common_prefix(i + 1, j + 1)         else:             return 0       # diagonally computing the subproblems     # to decrease memory dependency     def digonal_computation():                   # upper right triangle of the 2D array         for k in range(len(x)):                    yield from ((longest_common_prefix(i, j), i, j)                         for i, j in zip(range(k, -1, -1),                                     range(len(y) - 1, -1, -1)))                   # lower left triangle of the 2D array         for k in range(len(y)):                    yield from ((longest_common_prefix(i, j), i, j)                         for i, j in zip(range(k, -1, -1),                                     range(len(x) - 1, -1, -1)))       # returning the maximum of all the subproblems     return max(digonal_computation(), key=itemgetter(0), default=(0, 0, 0))   # Driver Code if __name__ == '__main__':     x: str = 'GeeksforGeeks'     y: str = 'GeeksQuiz'     length, i, j = longest_common_substring(x, y)     print(f'length: {length}, i: {i}, j: {j}')     print(f'x substring: {x[i: i + length]}')     print(f'y substring: {y[j: j + length]}')

Output

length: 5, i: 0, j: 0
x substring: Geeks
y substring: Geeks

Time Complexity
Space Complexity

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

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