Longest Palindromic Substring | Set 1
Given a string, find the longest substring which is palindrome.
For example,
Input: Given string :"forgeeksskeegfor", Output: "geeksskeeg" Input: Given string :"Geeks", Output: "ee"
Method 1: Brute Force.
Approach: The simple approach is to check each substring whether the substring is a palindrome or not. To do this first, run three nested loops, the outer two loops pick all substrings one by one by fixing the corner characters, the inner loop checks whether the picked substring is palindrome or not.
C++
// A C++ solution for longest palindrome #include <bits/stdc++.h> using namespace std; // Function to print a substring str[low..high] void printSubStr(string str, int low, int high) { for ( int i = low; i <= high; ++i) cout << str[i]; } // This function prints the // longest palindrome substring // It also returns the length // of the longest palindrome int longestPalSubstr(string str) { // get length of input string int n = str.size(); // All substrings of length 1 // are palindromes int maxLength = 1, start = 0; // Nested loop to mark start and end index for ( int i = 0; i < str.length(); i++) { for ( int j = i; j < str.length(); j++) { int flag = 1; // Check palindrome for ( int k = 0; k < (j - i + 1) / 2; k++) if (str[i + k] != str[j - k]) flag = 0; // Palindrome if (flag && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } cout << "Longest palindrome substring is: " ; printSubStr(str, start, start + maxLength - 1); // return length of LPS return maxLength; } // Driver Code int main() { string str = "forgeeksskeegfor" ; cout << "\nLength is: " << longestPalSubstr(str); return 0; } |
Java
// A Java solution for longest palindrome import java.util.*; class GFG{ // Function to print a subString str[low..high] static void printSubStr(String str, int low, int high) { for ( int i = low; i <= high; ++i) System.out.print(str.charAt(i)); } // This function prints the // longest palindrome subString // It also returns the length // of the longest palindrome static int longestPalSubstr(String str) { // get length of input String int n = str.length(); // All subStrings of length 1 // are palindromes int maxLength = 1 , start = 0 ; // Nested loop to mark start and end index for ( int i = 0 ; i < str.length(); i++) { for ( int j = i; j < str.length(); j++) { int flag = 1 ; // Check palindrome for ( int k = 0 ; k < (j - i + 1 ) / 2 ; k++) if (str.charAt(i + k) != str.charAt(j - k)) flag = 0 ; // Palindrome if (flag!= 0 && (j - i + 1 ) > maxLength) { start = i; maxLength = j - i + 1 ; } } } System.out.print( "Longest palindrome subString is: " ); printSubStr(str, start, start + maxLength - 1 ); // return length of LPS return maxLength; } // Driver Code public static void main(String[] args) { String str = "forgeeksskeegfor" ; System.out.print( "\nLength is: " + longestPalSubstr(str)); } } // This code is contributed by shikhasingrajput |
Python3
# A Python3 solution for longest palindrome # Function to pra subString str[low..high] def printSubStr( str , low, high): for i in range (low, high + 1 ): print ( str [i], end = "") # This function prints the # longest palindrome subString # It also returns the length # of the longest palindrome def longestPalSubstr( str ): # Get length of input String n = len ( str ) # All subStrings of length 1 # are palindromes maxLength = 1 start = 0 # Nested loop to mark start # and end index for i in range (n): for j in range (i, n): flag = 1 # Check palindrome for k in range ( 0 , ((j - i) / / 2 ) + 1 ): if ( str [i + k] ! = str [j - k]): flag = 0 # Palindrome if (flag ! = 0 and (j - i + 1 ) > maxLength): start = i maxLength = j - i + 1 print ( "Longest palindrome subString is: " , end = "") printSubStr( str , start, start + maxLength - 1 ) # Return length of LPS return maxLength # Driver Code if __name__ = = '__main__' : str = "forgeeksskeegfor" print ( "\nLength is: " , longestPalSubstr( str )) # This code is contributed by 29AjayKumar |
C#
// A C# solution for longest palindrome using System; class GFG{ // Function to print a subString str[low..high] static void printSubStr(String str, int low, int high) { for ( int i = low; i <= high; ++i) Console.Write(str[i]); } // This function prints the // longest palindrome subString // It also returns the length // of the longest palindrome static int longestPalSubstr(String str) { // get length of input String int n = str.Length; // All subStrings of length 1 // are palindromes int maxLength = 1, start = 0; // Nested loop to mark start and end index for ( int i = 0; i < str.Length; i++) { for ( int j = i; j < str.Length; j++) { int flag = 1; // Check palindrome for ( int k = 0; k < (j - i + 1) / 2; k++) if (str[i + k] != str[j - k]) flag = 0; // Palindrome if (flag!=0 && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } Console.Write( "longest palindrome subString is: " ); printSubStr(str, start, start + maxLength - 1); // return length of LPS return maxLength; } // Driver Code public static void Main(String[] args) { String str = "forgeeksskeegfor" ; Console.Write( "\nLength is: " + longestPalSubstr(str)); } } // This code is contributed by shikhasingrajput |
Javascript
<script> // A Javascript solution for longest palindrome // Function to print a subString str[low..high] function printSubStr(str,low,high) { for (let i = low; i <= high; ++i) document.write(str[i]); } // This function prints the // longest palindrome subString // It also returns the length // of the longest palindrome function longestPalSubstr(str) { // get length of input String let n = str.length; // All subStrings of length 1 // are palindromes let maxLength = 1, start = 0; // Nested loop to mark start and end index for (let i = 0; i < str.length; i++) { for (let j = i; j < str.length; j++) { let flag = 1; // Check palindrome for (let k = 0; k < (j - i + 1) / 2; k++) if (str[i + k] != str[j - k]) flag = 0; // Palindrome if (flag!=0 && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } document.write( "Longest palindrome subString is: " ); printSubStr(str, start, start + maxLength - 1); // return length of LPS return maxLength; } // Driver Code let str = "forgeeksskeegfor" ; document.write( "<br>Length is: " + longestPalSubstr(str)); // This code is contributed by rag2127 </script> |
Output:
Longest palindrome subString is: geeksskeeg Length is: 10
Complexity Analysis:
- Time complexity: O(n^3).
Three nested loops are needed to find the longest palindromic substring in this approach, so the time complexity is O(n^3). - Auxiliary complexity: O(1).
As no extra space is needed.
Method 2: Dynamic Programming.
Approach: The time complexity can be reduced by storing results of sub-problems. The idea is similar to this post.
- Maintain a boolean table[n][n] that is filled in bottom up manner.
- The value of table[i][j] is true, if the substring is palindrome, otherwise false.
- To calculate table[i][j], check the value of table[i+1][j-1], if the value is true and str[i] is same as str[j], then we make table[i][j] true.
- Otherwise, the value of table[i][j] is made false.
- We have to fill table previously for substring of length = 1 and length =2 because
as we are finding , if table[i+1][j-1] is true or false , so in case of
(i) length == 1 , lets say i=2 , j=2 and i+1,j-1 doesn’t lies between [i , j]
(ii) length == 2 ,lets say i=2 , j=3 and i+1,j-1 again doesn’t lies between [i , j].
Below is the implementation of the above approach:
C++
// A C++ dynamic programming // solution for longest palindrome #include <bits/stdc++.h> using namespace std; // Function to print a substring // str[low..high] void printSubStr( string str, int low, int high) { for ( int i = low; i <= high; ++i) cout << str[i]; } // This function prints the // longest palindrome substring // It also returns the length of // the longest palindrome int longestPalSubstr(string str) { // get length of input string int n = str.size(); // table[i][j] will be false if substring // str[i..j] is not palindrome. // Else table[i][j] will be true bool table[n][n]; memset (table, 0, sizeof (table)); // All substrings of length 1 // are palindromes int maxLength = 1; for ( int i = 0; i < n; ++i) table[i][i] = true ; // check for sub-string of length 2. int start = 0; for ( int i = 0; i < n - 1; ++i) { if (str[i] == str[i + 1]) { table[i][i + 1] = true ; start = i; maxLength = 2; } } // Check for lengths greater than 2. // k is length of substring for ( int k = 3; k <= n; ++k) { // Fix the starting index for ( int i = 0; i < n - k + 1; ++i) { // Get the ending index of substring from // starting index i and length k int j = i + k - 1; // checking for sub-string from ith index to // jth index iff str[i+1] to str[j-1] is a // palindrome if (table[i + 1][j - 1] && str[i] == str[j]) { table[i][j] = true ; if (k > maxLength) { start = i; maxLength = k; } } } } cout << "Longest palindrome substring is: " ; printSubStr(str, start, start + maxLength - 1); // return length of LPS return maxLength; } // Driver Code int main() { string str = "forgeeksskeegfor" ; cout << "\nLength is: " << longestPalSubstr(str); return 0; } |
Java
// Java Solution public class LongestPalinSubstring { // A utility function to print // a substring str[low..high] static void printSubStr( String str, int low, int high) { System.out.println( str.substring( low, high + 1 )); } // This function prints the longest // palindrome substring of str[]. // It also returns the length of the // longest palindrome static int longestPalSubstr(String str) { // get length of input string int n = str.length(); // table[i][j] will be false if // substring str[i..j] is not palindrome. // Else table[i][j] will be true boolean table[][] = new boolean [n][n]; // All substrings of length 1 are palindromes int maxLength = 1 ; for ( int i = 0 ; i < n; ++i) table[i][i] = true ; // check for sub-string of length 2. int start = 0 ; for ( int i = 0 ; i < n - 1 ; ++i) { if (str.charAt(i) == str.charAt(i + 1 )) { table[i][i + 1 ] = true ; start = i; maxLength = 2 ; } } // Check for lengths greater than 2. // k is length of substring for ( int k = 3 ; k <= n; ++k) { // Fix the starting index for ( int i = 0 ; i < n - k + 1 ; ++i) { // Get the ending index of substring from // starting index i and length k int j = i + k - 1 ; // checking for sub-string from ith index to // jth index iff str.charAt(i+1) to // str.charAt(j-1) is a palindrome if (table[i + 1 ][j - 1 ] && str.charAt(i) == str.charAt(j)) { table[i][j] = true ; if (k > maxLength) { start = i; maxLength = k; } } } } System.out.print( "Longest palindrome substring is; " ); printSubStr(str, start, start + maxLength - 1 ); // return length of LPS return maxLength; } // Driver program to test above functions public static void main(String[] args) { String str = "forgeeksskeegfor" ; System.out.println( "Length is: " + longestPalSubstr(str)); } } // This code is contributed by Sumit Ghosh |
Python
# Python program import sys # A utility function to print a # substring str[low..high] def printSubStr(st, low, high) : sys.stdout.write(st[low : high + 1 ]) sys.stdout.flush() return '' # This function prints the longest palindrome # substring of st[]. It also returns the length # of the longest palindrome def longestPalSubstr(st) : n = len (st) # get length of input string # table[i][j] will be false if substring # str[i..j] is not palindrome. Else # table[i][j] will be true table = [[ 0 for x in range (n)] for y in range (n)] # All substrings of length 1 are # palindromes maxLength = 1 i = 0 while (i < n) : table[i][i] = True i = i + 1 # check for sub-string of length 2. start = 0 i = 0 while i < n - 1 : if (st[i] = = st[i + 1 ]) : table[i][i + 1 ] = True start = i maxLength = 2 i = i + 1 # Check for lengths greater than 2. # k is length of substring k = 3 while k < = n : # Fix the starting index i = 0 while i < (n - k + 1 ) : # Get the ending index of # substring from starting # index i and length k j = i + k - 1 # checking for sub-string from # ith index to jth index iff # st[i + 1] to st[(j-1)] is a # palindrome if (table[i + 1 ][j - 1 ] and st[i] = = st[j]) : table[i][j] = True if (k > maxLength) : start = i maxLength = k i = i + 1 k = k + 1 print "Longest palindrome substring is: " , printSubStr(st, start, start + maxLength - 1 ) return maxLength # return length of LPS # Driver program to test above functions st = "forgeeksskeegfor" l = longestPalSubstr(st) print "Length is:" , l # This code is contributed by Nikita Tiwari. |
C#
// C# Solution using System; class GFG { // A utility function to print a // substring str[low...( high - (low+1))] static void printSubStr( string str, int low, int high) { Console.WriteLine(str.Substring(low, high - low + 1)); } // This function prints the longest // palindrome substring of str[]. // It also returns the length of the // longest palindrome static int longestPalSubstr( string str) { // Get length of input string int n = str.Length; // Table[i, j] will be false if substring // str[i..j] is not palindrome. Else // table[i, j] will be true bool [, ] table = new bool [n, n]; // All substrings of length 1 are palindromes int maxLength = 1; for ( int i = 0; i < n; ++i) table[i, i] = true ; // Check for sub-string of length 2. int start = 0; for ( int i = 0; i < n - 1; ++i) { if (str[i] == str[i + 1]) { table[i, i + 1] = true ; start = i; maxLength = 2; } } // Check for lengths greater than 2. // k is length of substring for ( int k = 3; k <= n; ++k) { // Fix the starting index for ( int i = 0; i < n - k + 1; ++i) { // Get the ending index of substring from // starting index i and length k int j = i + k - 1; // Checking for sub-string from ith index // to jth index iff str.charAt(i+1) to // str.charAt(j-1) is a palindrome if (table[i + 1, j - 1] && str[i] == str[j]) { table[i, j] = true ; if (k > maxLength) { start = i; maxLength = k; } } } } Console.Write( "Longest palindrome substring is: " ); printSubStr(str, start, start + maxLength - 1); // Return length of LPS return maxLength; } // Driver code public static void Main( string [] args) { string str = "forgeeksskeegfor" ; Console.WriteLine( "Length is: " + longestPalSubstr(str)); } } // This code is contributed by SoumikMondal |
Javascript
<script> // Javascript Solution // A utility function to print // a substring str[low..high] function printSubStr(str,low,high) { document.write( str.substring( low, high + 1)+ "<br>" ); } // This function prints the longest // palindrome substring of str[]. // It also returns the length of the // longest palindrome function longestPalSubstr(str) { // get length of input string let n = str.length; // table[i][j] will be false if // substring str[i..j] is not palindrome. // Else table[i][j] will be true let table = new Array(n); for (let i = 0; i < n; i++) { table[i] = new Array(n); } // All substrings of length 1 are palindromes let maxLength = 1; for (let i = 0; i < n; ++i) table[i][i] = true ; // check for sub-string of length 2. let start = 0; for (let i = 0; i < n - 1; ++i) { if (str[i] == str[i + 1]) { table[i][i + 1] = true ; start = i; maxLength = 2; } } // Check for lengths greater than 2. // k is length of substring for (let k = 3; k <= n; ++k) { // Fix the starting index for (let i = 0; i < n - k + 1; ++i) { // Get the ending index of substring from // starting index i and length k let j = i + k - 1; // checking for sub-string from ith index to // jth index iff str.charAt(i+1) to // str.charAt(j-1) is a palindrome if (table[i + 1][j - 1] && str[i] == str[j]) { table[i][j] = true ; if (k > maxLength) { start = i; maxLength = k; } } } } document.write( "Longest palindrome substring is; " ); printSubStr(str, start, start + maxLength - 1); // return length of LPS return maxLength; } // Driver program to test above functions let str = "forgeeksskeegfor" ; document.write( "Length is: " + longestPalSubstr(str)); // This code is contributed by avanitrachhadiya2155 </script> |
Output:
Longest palindrome substring is: geeksskeeg Length is: 10
Complexity Analysis:
- Time complexity: O(n^2).
Two nested traversals are needed. - Auxiliary Space: O(n^2).
Matrix of size n*n is needed to store the dp array.
A better space complexity approach can be found in Set-2.
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