Longest Common Substring | DP-29
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 : 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 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. |
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:GeeksforGeeks.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:GeeksforGeeks.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> |
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 <bits/stdc++.h> 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
<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:GeeksforGeeks.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> |
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 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 |
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> |
4
Maximum Space Optimization:
- In this method, we will use recursion to find the longest prefix of all the possible substrings.
- Let
gives the length of the longest common suffix starting from indices i, j of strings X, Y respectively.
- Then the function can be defined as :
- 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.
- Consider the following table where we memorize the solutions:
0 | 1 | 2 | 3 | 4 | |
0 | |||||
1 | |||||
2 | |||||
3 |
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]}' ) |
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.