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Print the longest common substring

Last Updated : 15 Jul, 2022
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Given two strings ‘X’ and ‘Y’, print the length of the longest common substring. If two or more substrings have the same value for the longest common substring, then print any one of them.

Examples: 

Input :  X = "GeeksforGeeks", 
         Y = "GeeksQuiz"
Output : Geeks

Input : X = "zxabcdezy", 
        Y = "yzabcdezx"
Output : abcdez

We have discussed a solution to find the length of the longest common string. In this post, we have discussed printing common string is discussed.

Naive Approach: Let strings X and Y be the lengths m and n respectively. Generate all possible substrings of X which requires a time complexity of O(m2) and search each substring in the string Y which can be achieved in O(n) time complexity using the KMP algorithm. Overall time complexity will be O(n * m2).

Efficient Approach: It is based on the dynamic programming implementation explained in this post. The longest suffix matrix LCSuff[][] is build up and the index of the cell having the maximum value is tracked. Let that index be represented by (row, col) pair. Now the final longest common substring is built with the help of that index by diagonally traversing up the LCSuff[][] matrix until LCSuff[row][col] != 0 and during the iteration obtaining the characters either from X[row-1] or Y[col-1] and adding them from right to left in the resultant common string. 

Implementation:

C++




// C++ implementation to print the longest common substring
#include <iostream>
#include <stdlib.h>
#include <string.h>
 
using namespace std;
 
/* function to find and print the longest common
   substring of X[0..m-1] and Y[0..n-1] */
void printLCSubStr(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 LCSuff[m + 1][n + 1];
 
    // To store length of the longest common substring
    int len = 0;
 
    // To store the index of the cell which contains the
    // maximum value. This cell's index helps in building
    // up the longest common substring from right to left.
    int row, col;
 
    /* 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;
                if (len < LCSuff[i][j]) {
                    len = LCSuff[i][j];
                    row = i;
                    col = j;
                }
            }
            else
                LCSuff[i][j] = 0;
        }
    }
 
    // if true, then no common substring exists
    if (len == 0) {
        cout << "No Common Substring";
        return;
    }
 
    // allocate space for the longest common substring
    char* resultStr = (char*)malloc((len + 1) * sizeof(char));
 
    // traverse up diagonally form the (row, col) cell
    // until LCSuff[row][col] != 0
    while (LCSuff[row][col] != 0) {
        resultStr[--len] = X[row - 1]; // or Y[col-1]
 
        // move diagonally up to previous cell
        row--;
        col--;
    }
 
    // required longest common substring
    cout << resultStr;
}
 
/* 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);
 
    printLCSubStr(X, Y, m, n);
    return 0;
}


Java




// Java implementation to print the longest common substring
public class Longest_common_substr {
 
    /* function to find and print the longest common
       substring of X[0..m-1] and Y[0..n-1] */
    static void printLCSubStr(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[][] LCSuff = new int[m + 1][n + 1];
 
        // To store length of the longest common substring
        int len = 0;
 
        // To store the index of the cell which contains the
        // maximum value. This cell's index helps in building
        // up the longest common substring from right to left.
        int row = 0, col = 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)
                    LCSuff[i][j] = 0;
 
                else if (X.charAt(i - 1) == Y.charAt(j - 1)) {
                    LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1;
                    if (len < LCSuff[i][j]) {
                        len = LCSuff[i][j];
                        row = i;
                        col = j;
                    }
                }
                else
                    LCSuff[i][j] = 0;
            }
        }
 
        // if true, then no common substring exists
        if (len == 0) {
            System.out.println("No Common Substring");
            return;
        }
 
        // allocate space for the longest common substring
        String resultStr = "";
 
        // traverse up diagonally form the (row, col) cell
        // until LCSuff[row][col] != 0
        while (LCSuff[row][col] != 0) {
            resultStr = X.charAt(row - 1) + resultStr; // or Y[col-1]
            --len;
 
            // move diagonally up to previous cell
            row--;
            col--;
        }
 
        // required longest common substring
        System.out.println(resultStr);
    }
 
    /* 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();
 
        printLCSubStr(X, Y, m, n);
    }
}
// This code is contributed by Sumit Ghosh


Python3





C#





Javascript





Output: 

Site:Geeks

 

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

Space Optimized Approach: The auxiliary space used by the solution above is O(m*n), where m and n are lengths of string X and Y. The space used by the above solution can be reduced to O(2*n). A variable end is used to store the ending point of the longest common substring in string X and variable maxlen is used to store the length of the longest common substring. 

Suppose we are at DP state when the length of X is i and length of Y is j, the result of which is stored in len[i][j]. 
Now if X[i-1] == Y[j-1], then len[i][j] = 1 + len[i-1][j-1], that is result of current row in matrix len[][] depends on values from previous row. Hence, the required length of the longest common substring can be obtained by maintaining values of two consecutive rows only, thereby reducing space requirements to O(2*n). 

To print the longest common substring, we use a variable end. When len[i][j] is calculated, it is compared with maxlen. If maxlen is less than len[i][j], then end is updated to i-1 to show that longest common substring ends at index i-1 in X and maxlen is updated to len[i][j]. The longest common substring then is from index end – maxlen + 1 to index end in X. 
A variable currRow is used to represent that either row 0 or row 1 of len[2][n] matrix is currently used to find the length. Initially, row 0 is used as the current row for the case when the length of string X is zero. At the end of each iteration, the current row is made the previous row and the previous row is made the new current row.

Given below is the implementation of the above approach: 

C++




// Space optimized CPP implementation to print
// longest common substring.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find longest common substring.
string 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;
 
    // Variable to store ending point of
    // longest common substring in X.
    int end;
 
    // Matrix to store result of two
    // consecutive rows at a time.
    int len[2][n + 1];
 
    // 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;
                if (len[currRow][j] > result) {
                    result = len[currRow][j];
                    end = i - 1;
                }
            }
            else {
                len[currRow][j] = 0;
            }
        }
 
        // Make current row as previous row and
        // previous row as new current row.
        currRow = 1 - currRow;
    }
 
    // If there is no common substring, print -1.
    if (result == 0) {
        return "-1";
    }
 
    // Longest common substring is from index
    // end - result + 1 to index end in X.
    return X.substr(end - result + 1, result);
}
// Driver Code
int main()
{
    string X = "GeeksforGeeks";
    string Y = "GeeksQuiz";
    // function call
    cout << LCSubStr(X, Y);
    return 0;
}


Java




// Space optimized Java implementation to print
// longest common substring.
 
public class GFG {
 
// Function to find longest common substring.
    static String 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;
 
        // Variable to store ending point of
        // longest common subString in X.
        int end = 0;
 
        // Matrix to store result of two
        // consecutive rows at a time.
        int len[][] = new int[2][m];
 
        // 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;
                    if (len[currRow][j] > result) {
                        result = len[currRow][j];
                        end = i - 1;
                    }
                } else {
                    len[currRow][j] = 0;
                }
            }
 
            // Make current row as previous row and
            // previous row as new current row.
            currRow = 1 - currRow;
        }
 
        // If there is no common subString, print -1.
        if (result == 0) {
            return "-1";
        }
 
        // Longest common subString is from index
        // end - result + 1 to index end in X.
        return X.substring(end - result + 1, result);
    }
     
    // Driver Code
    public static void main(String[] args) {
        String X = "GeeksforGeeks";
        String Y = "GeeksQuiz";
        // function call
        System.out.println(LCSubStr(X, Y));
 
    }
 
}
// This code is contributed by PrinciRaj1992


Python3




# Space optimized Python3 implementation to
# print longest common substring.
 
# 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
  
    # Variable to store ending point of
    # longest common substring in X.
    end = 0
  
    # Matrix to store result of two
    # consecutive rows at a time.
    length = [[0 for j in range(m)]
                 for i in range(2)]
  
    # Variable to represent which row of
    # matrix is current row.
    currRow = 0
  
    # For a particular value of i and j,
    # length[currRow][j] stores length
    # of longest common substring in
    # string X[0..i] and Y[0..j].
    for i in range(0, m + 1):
        for j in range(0, n + 1):
            if (i == 0 or j == 0):
                length[currRow][j] = 0
             
            elif (X[i - 1] == Y[j - 1]):
                length[currRow][j] = length[1 - currRow][j - 1] + 1
                 
                if (length[currRow][j] > result):
                    result = length[currRow][j]
                    end = i - 1
            else:
                length[currRow][j] = 0
  
        # Make current row as previous row and
        # previous row as new current row.
        currRow = 1 - currRow
  
    # If there is no common substring, print -1.
    if (result == 0):
        return "-1"
  
    # Longest common substring is from index
    # end - result + 1 to index end in X.
    return X[end - result + 1 : end + 1]
     
# Driver code
if __name__=="__main__":
     
    X = "GeeksforGeeks"
    Y = "GeeksQuiz"
     
    # Function call
    print(LCSubStr(X, Y))
 
# This code is contributed by rutvik_56


C#





Javascript




<script>
// Space optimized javascript implementation to print
// longest common substring.   
     
    // Function to find longest common substring.
    function LCSubStr(X,Y)
    {
        // Find length of both the strings.
    let m = X.length;
    let n = Y.length;
  
    // Variable to store length of longest
    // common substring.
    let result = 0;
  
    // Variable to store ending point of
    // longest common substring in X.
    let end;
  
    // Matrix to store result of two
    // consecutive rows at a time.
    let len= new Array(2);
    for(let i=0;i<len.length;i++)
    {
        len[i]=new Array(n);
        for(let j=0;j<n;j++)
        {
            len[i][j]=0;
        }
    }
  
    // Variable to represent which row of
    // matrix is current row.
    let 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 (let i = 0; i <= m; i++) {
        for (let 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;
                if (len[currRow][j] > result) {
                    result = len[currRow][j];
                    end = i - 1;
                }
            }
            else {
                len[currRow][j] = 0;
            }
        }
  
        // Make current row as previous row and
        // previous row as new current row.
        currRow = 1 - currRow;
    }
  
    // If there is no common substring, print -1.
    if (result == 0) {
        return "-1";
    }
  
    // Longest common substring is from index
    // end - result + 1 to index end in X.
    return X.substr(end - result + 1, result);
    }
     
    // Driver Code
    let X = "GeeksforGeeks";
    let Y = "GeeksQuiz";
    // function call
    document.write(LCSubStr(X, Y));
     
     
    // This code is contributed by avanitrachhadiya2155
</script>


Output: 

Geeks

 

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

This approach has been suggested by nik1996. 



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