Given two strings ‘X’ and ‘Y’, find the length of the longest common substring.
Examples :
Input : X = “neveropen”, y = “GeeksQuiz”
Output : 5
Explanation:
The longest common substring is “Geeks” and is of length 5.Input : X = “abcdxyz”, y = “xyzabcd”
Output : 4
Explanation:
The longest common substring is “abcd” and is of length 4.Input : X = “zxabcdezy”, y = “yzabcdezx”
Output : 6
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 <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 codeint main(){ char X[] = "OldSite:neveropen.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 Programmingimport java.io.*;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:neveropen.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 CodeX = 'OldSite:neveropen.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:neveropen.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 $X = "OldSite:neveropen.org";$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?> |
Javascript
<script>// JavaScript implementation of// finding length of longest// Common substring using // Dynamic Programming /* 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. // 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 var LCStuff = Array(m + 1).fill().map(()=>Array(n + 1).fill(0)); // To store length of the longest // common substring var result = 0; // 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) 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 var X = "OldSite:neveropen.org"; var Y = "NewSite:GeeksQuiz.com"; var m = X.length; var n = Y.length; document.write("Length of Longest Common Substring is " + LCSubStr(X, Y, m, n));// This code contributed by Rajput-Ji</script> |
Output
Length of Longest Common Substring is 10
Time Complexity: O(m*n)
Auxiliary Space: O(m*n), since m*n extra space has been taken.
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 <bits/stdc++.h>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[2][m + 1]; int res = 0; dp[0][0] = 0; dp[1][0] = 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:neveropen.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 approachimport java.io.*;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:neveropen.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:neveropen.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:neveropen.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
<script>// JavaScript implementation of the above approach // Function to find the length of the // longest LCS function LCSubStr(s, t, n, m) { // Create DP table var dp = Array(2).fill().map(()=>Array(m+ 1).fill(0)); var res = 0; for(var i = 1; i <= n; i++) { for(var 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 var X = "OldSite:neveropen.org"; var Y = "NewSite:GeeksQuiz.com"; var m = X.length; var n = Y.length; // Function call document.write(LCSubStr(X, Y, m, n));// This code is contributed by shivanisinghss2110</script> |
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 <iostream>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 recursionimport java.io.*;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
<?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?> |
Javascript
<script> // Javascript program using to find length of the // longest common substring recursion let X, Y; // Returns length of function f // or longest common substring // of X[0..m-1] and Y[0..n-1] function lcs(i, j, 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; } let n, m; X = "abcdxyz"; Y = "xyzabcd"; n = X.length; m = Y.length; document.write(lcs(n, m, 0)); // This code is contributed by divyeshrabadiya07.</script> |
Output
4
Time complexity: O(2^max(m,n)) as the function is doing two recursive calls – lcs(i, j-1, 0) and lcs(i-1, j, 0) when characters at X[i-1] != Y[j-1]. So it will give a worst case time complexity as 2^N, where N = max(m, n), m and n is the length of X and Y string.
Auxiliary Space: O(1): as the function call is not using any extra space (function is just using a recursive call stack which we generally doesn’t consider in auxiliary space).
Maximum Space Optimization:Ad
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