# Longest Common Substring | DP-29

• Difficulty Level : Medium
• Last Updated : 15 Nov, 2021

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

Examples :

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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. 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 codeint 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 Programmingclass 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 CodeX = '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 Programmingusing 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.

## PHP

 

## 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 LCSint LCSubStr(string s, string t, int n, int m){       // Create DP table    int dp[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 Codeint 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[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 LCSdef 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 CodeX = "OldSite:GeeksforGeeks.org"Y = "NewSite:GeeksQuiz.com"m = len(X)n = len(Y) # Function callprint(LCSubStr(X,Y,m,n)) # This code is contributed by avanitrachhadiya2155

## C#

 // C# implementation of the above approachusing 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 codeint 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 codeif __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// recursionusing 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

 

## 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 approachfrom functools import lru_cachefrom 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 Codeif __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 References: http://en.wikipedia.org/wiki/Longest_common_substring_problem
The longest substring can also be solved in O(n+m) time using the Suffix Tree. We will be covering the Suffix Tree-based solution in a separate post.
Exercise: The above solution prints only the length of the longest common substring. Extend the solution to print the substring also.