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Longest Palindromic Substring | Set 2

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  • Difficulty Level : Medium
  • Last Updated : 30 Jun, 2022
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Given a string, find the longest substring which is a palindrome. 

Examples: 

Input: Given string :"forgeeksskeegfor", 
Output: "geeksskeeg".

Input: Given string :"Geeks", 
Output: "ee". 

Common mistake : Wrong Approach:   

Some people will be tempted to come up with a O(n) time complexity quick solution, which is unfortunately flawed (however can be corrected easily):

 (i)Reverse S and  and store it in S’. 
 (ii)Find the longest common substring between S and S’  which must also be the longest palindromic substring.

This seemed to work, let’s see some examples below.

For example, S = “caba” then  S’ = “abac”.

The longest common substring between S and S’ is “aba”, which is the answer.

Let’s try another example: S = “abacdfgdcaba” then S’ = “abacdgfdcaba”.

The longest common substring between S and S’ is “abacd”. Clearly, this is not a valid palindrome.

Correct Approach:

We could see that the longest common substring method fails when there exists a reversed copy of a non-palindromic substring in some other part of S. To rectify this, each time we find a longest common substring candidate, we check if the substring’s indices are the same as the reversed substring’s original indices. If it is, then we attempt to update the longest palindrome found so far; if not, we skip this and find the next candidate. This gives us an O(n^2) Dynamic Programming solution which uses O(n^2) space (which could be improved to use O(n) space).

Dyanamic programming Approach: Dynamic programming solution is already discussed here in the previous post. The time complexity of the Dynamic Programming based solution is O(n^2) and it requires O(n^2) extra space. We can find the longest palindrome substring( LPS ) in (n^2) time with O(1) extra space. 

The algorithm below is very simple and easy to understand. The idea is to Fix a center and expand in both directions for longer palindromes and keep track of the longest palindrome seen so far.

ALGO:

  1. Maintain a variable ‘ maxLength = 1 ‘ (for storing LPS length) and ‘ start =0 ‘ (for storing starting index of LPS ).
  2. The idea is very simple, we will traverse through the entire string with i=0 to i<(length of string).
    1. while traversing, initialize ‘lowand ‘highpointer such that low= i-1 and high= i+1.
    2. keep incrementing ‘high’ until str[high]==str[i] .
    3. similarly keep decrementing ‘low’ until str[low]==str[i].
    4. finally we will keep incrementing ‘high’ and decrementing ‘low’ until str[low]==str[high].
    5. calculate length=high-low-1, if length > maxLength then maxLength = length and start = low+1 .
  3. Print the LPS and return maxLength.
     

C++




// A O(n^2) time and O(1) space program to
// find the longest palindromic substring
// easy to understand as compared to previous version.
#include <bits/stdc++.h>
using namespace std;
 
// A utility function to print
// a substring str[low..high]
// This function prints the
// longest palindrome substring (LPS)
// of str[]. It also returns the
// length of the longest palindrome
int longestPalSubstr(string str)
{
    int n = str.size(); // calculating size of string
    if (n < 2)
        return n; // if string is empty then size will be 0.
                  // if n==1 then, answer will be 1(single
                  // character will always palindrome)
 
    int maxLength = 1, start = 0;
    int low, high;
    for (int i = 0; i < n; i++) {
        low = i - 1;
        high = i + 1;
        while (high < n
               && str[high] == str[i]) // increment 'high'
            high++;
 
        while (low >= 0
               && str[low] == str[i]) // decrement 'low'
            low--;
 
        while (low >= 0 && high < n
               && str[low] == str[high]) {
            low--;
            high++;
        }
 
        int length = high - low - 1;
        if (maxLength < length) {
            maxLength = length;
            start = low + 1;
        }
    }
 
    cout << "Longest palindrome substring is: ";
    cout << str.substr(start, maxLength);
    return maxLength;
}
 
// Driver program to test above functions
int main()
{
    string str = "forgeeksskeegfor";
    cout << "\nLength is: " << longestPalSubstr(str)
         << endl;
    return 0;
}


C




// A O(n^2) time and O(1) space
// program to find the longest
// palindromic substring
#include <stdio.h>
#include <string.h>
 
// A utility function to print
// a substring str[low..high]
void printSubStr(char* str, int low, int high)
{
    for (int i = low; i <= high; ++i)
        printf("%c", str[i]);
}
 
// This function prints the longest
// palindrome substring (LPS)
// of str[]. It also returns the
// length of the longest palindrome
int longestPalSubstr(char* str)
{
    int n = strlen(str); // calculating size of string
    if (n < 2)
        return n; // if string is empty then size will be 0.
                  // if n==1 then, answer will be 1(single
                  // character will always palindrome)
 
    int maxLength = 1, start = 0;
    int low, high;
    for (int i = 0; i < n; i++) {
        low = i - 1;
        high = i + 1;
        while (high < n
               && str[high] == str[i]) // increment 'high'
            high++;
 
        while (low >= 0
               && str[low] == str[i]) // decrement 'low'
            low--;
 
        while (low >= 0 && high < n
               && str[low] == str[high]) {
            low--; // decrement low
            high++; // increment high
        }
 
        int length = high - low - 1;
        if (maxLength < length) {
            maxLength = length;
            start = low + 1;
        }
    }
    printf("Longest palindrome substring is: ");
    printSubStr(str, start, start + maxLength - 1);
    return maxLength;
}
 
// Driver program to test above functions
int main()
{
    char str[] = "forgeeksskeegfor";
    printf("\nLength is: %d", longestPalSubstr(str));
    return 0;
}


Java




// Java implementation of O(n^2)
// time and O(1) space method
// to find the longest palindromic substring
public class LongestPalinSubstring {
   
    // This function prints the
    // longest palindrome substring
    // (LPS) of str[]. It also
    // returns the length of the
    // longest palindrome
    static int longestPalSubstr(String str)
    {
        int n = str.length(); // calculcharAting size of string
        if (n < 2)
            return n; // if string is empty then size will be 0.
                  // if n==1 then, answer will be 1(single
                  // character will always palindrome)
 
        int maxLength = 1,start=0;
        int low, high;
        for (int i = 0; i < n; i++) {
            low = i - 1;
            high = i + 1;
            while ( high < n && str.charAt(high) == str.charAt(i)) //increment 'high'                                  
                high++;
       
            while ( low >= 0 && str.charAt(low) == str.charAt(i)) // decrement 'low'                   
                low--;
       
            while (low >= 0 && high < n && str.charAt(low) == str.charAt(high) ){
                  low--;
                high++;
        }
 
        int length = high - low - 1;
        if (maxLength < length){
            maxLength = length;
            start=low+1;
        }
    }   
    System.out.print("Longest palindrome substring is: ");
    System.out.println(str.substring(start, start + maxLength ));
    return maxLength;
       
 }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
 
        String str = "forgeeksskeegfor";
        System.out.println("Length is: "
                           + longestPalSubstr(str));
    }
}


Python3




# A O(n ^ 2) time and O(1) space program to find the
# longest palindromic substring
 
# This function prints the longest palindrome substring (LPS)
# of str[]. It also returns the length of the longest palindrome
 
 
def longestPalSubstr(string):
    n = len(string) # calculating size of string
    if (n < 2):
        return n # if string is empty then size will be 0.
                  # if n==1 then, answer will be 1(single
                  # character will always palindrome)
    start=0
    maxLength = 1
    for i in range(n):
        low = i - 1
        high = i + 1
        while (high < n and string[high] == string[i] ):                              
            high=high+1
       
        while (low >= 0 and string[low] == string[i] ):                
            low=low-1
       
        while (low >= 0 and high < n and string[low] == string[high] ):
          low=low-1
          high=high+1
         
     
        length = high - low - 1
        if (maxLength < length):
            maxLength = length
            start=low+1
             
    print ("Longest palindrome substring is:",end=" ")
    print (string[start:start + maxLength])
     
    return maxLength
     
# Driver program to test above functions
string = ("forgeeksskeegfor")
print("Length is: " + str(longestPalSubstr(string)))


C#




// C# implementation of O(n^2) time
// and O(1) space method to find the
// longest palindromic substring
using System;
 
class GFG {
 
    // This function prints the longest
    // palindrome substring (LPS) of str[].
    // It also returns the length of the
    // longest palindrome
    public static int longestPalSubstr(string str)
    {
        int n = str.Length; // calculcharAting size of string
        if (n < 2)
            return n; // if string is empty then size will be 0.
                  // if n==1 then, answer will be 1(single
                  // character will always palindrome)
 
        int maxLength = 1,start=0;
        int low, high;
        for (int i = 0; i < n; i++) {
            low = i - 1;
            high = i + 1;
            while ( high < n && str[high] == str[i] ) //increment 'high'                                  
                high++;
       
            while ( low >= 0 && str[low] == str[i]) // decrement 'low'                   
                low--;
       
            while (low >= 0 && high < n && str[low] == str[high] ){
                  low--;
                high++;
        }
 
        int length = high - low - 1;
        if (maxLength < length){
            maxLength = length;
            start=low+1;
        }
    }   
    Console.Write("Longest palindrome substring is: ");
    Console.WriteLine(str.Substring(start, maxLength));
    return maxLength;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        string str = "forgeeksskeegfor";
        Console.WriteLine("Length is: "
                          + longestPalSubstr(str));
    }
}


Javascript




<script>
 
// A O(n^2) time and O(1) space program to
// find the longest palindromic substring
// easy to understand as compared to previous version.
 
// A utility function to print
// a substring str[low..high]
// This function prints the
// longest palindrome substring (LPS)
// of str[]. It also returns the
// length of the longest palindrome
function longestPalSubstr(str)
{
    let n = str.length; // calculating size of string
    if (n < 2)
        return n; // if string is empty then size will be 0.
                // if n==1 then, answer will be 1(single
                // character will always palindrome)
 
    let maxLength = 1,start=0;
    let low, high;
    for (let i = 0; i < n; i++) {
        low = i - 1;
        high = i + 1;
        while ( high < n && str[high] == str[i]) //increment 'high'                               
            high++;
     
        while ( low >= 0 && str[low] == str[i]) // decrement 'low'               
            low--;
     
        while (low >= 0 && high < n && str[low] == str[high]){
            low--;
            high++;
        }
 
        let length = high - low - 1;
        if (maxLength < length) {
            maxLength = length;
            start=low+1;
        }
    }
     
    document.write("Longest palindrome substring is: ");
    document.write(str.substring(start,maxLength+start));
    return maxLength;
}
 
// Driver program to test above functions
 
let str = "forgeeksskeegfor";
document.write("</br>","Length is: " + longestPalSubstr(str),"</br>");
 
 
</script>


Output

Longest palindrome substring is: geeksskeeg
Length is: 10

Complexity Analysis:  

  • Time complexity: O(n^2), where n is the length of the input string. 
    Outer Loop that traverses through the entire string, and Inner Loop that is used to expand from i .
  • Auxiliary Space: O(1). 
    No extra space is needed.

The Above approach is a cleaner way.

The function implementation to print the LPS and return the maxLength is given below:

C++




// A O(n^2) time and O(1) space program to
// find the longest palindromic substring
// easy to understand as compared to previous version.
 
#include <bits/stdc++.h>
using namespace std;
 
int maxLength; // variables to store and
string res; // update  maxLength and res
 
// A utility function to get the longest palindrome
// starting and expanding out from given center indices
void cSubUtil(string& s, int l, int r)
{
    // check if the indices lie in the range of string
    // and also if it is palindrome
    while (l >= 0 && r < s.length() && s[l] == s[r]) {
        // expand the boundary
        l--;
        r++;
    }
    // if it's length is greater than maxLength update
    // maxLength and res
    if (r - l - 1 >= maxLength) {
        res = s.substr(l + 1, r - l - 1);
        maxLength = r - l - 1;
    }
    return;
}
// A function which takes a string prints the LPS and
// returns the length of LPS
int longestPalSubstr(string str)
{
    res = "";
    maxLength = 1;
    // for every index in the string check palindromes
    // starting from that index
    for (int i = 0; i < str.length(); i++) {
        // check for odd length palindromes
        cSubUtil(str, i, i);
        // check for even length palindromes
        cSubUtil(str, i, i + 1);
    }
 
    cout << "Longest palindrome substring is: ";
    cout << res << "\n";
 
    return maxLength;
}
 
// Driver program to test above functions
int main()
{
    string str = "forgeeksskeegfor";
    cout << "\nLength is: " << longestPalSubstr(str)
         << endl;
    return 0;
}


Java




// Java implementation of O(n^2)
// time and O(1) space method
// to find the longest palindromic substring
 
public class LongestPalinSubstring {
 
    static int maxLength; // variables to store and
    static String res; // update  maxLength and res
 
    // A utility function to get the longest palindrome
    // starting and expanding out from given center indices
    static void cSubUtil(String s, int l, int r)
    {
        // check if the indices lie in the range of string
        // and also if it is palindrome
        while (l >= 0 && r < s.length()
               && s.charAt(l) == s.charAt(r)) {
            // expand the boundary
            l--;
            r++;
        }
        // if it's length is greater than maxLength update
        // maxLength and res
        if (r - l - 1 >= maxLength) {
            res = s.substring(l + 1, r);
            maxLength = r - l - 1;
        }
        return;
    }
    // A function which takes a string prints the LPS and
    // returns the length of LPS
    static int longestPalSubstr(String str)
    {
        res = "";
        maxLength = 1;
        // for every index in the string check palindromes
        // starting from that index
        for (int i = 0; i < str.length(); i++) {
            // check for odd length palindromes
            cSubUtil(str, i, i);
            // check for even length palindromes
            cSubUtil(str, i, i + 1);
        }
        System.out.print(
            "Longest palindrome substring is: ");
        System.out.println(res);
        return maxLength;
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
 
        String str = "forgeeksskeegfor";
        System.out.println("Length is: "
                           + longestPalSubstr(str));
    }
}


C#




// C# implementation of O(n^2) time
// and O(1) space method to find the
// longest palindromic substring
using System;
 
class GFG {
 
    static int maxLength; // variables to store and
    static string res; // update  maxLength and res
 
    // A utility function to get the longest palindrome
    // starting and expanding out from given center indices
    public static void cSubUtil(string s, int l, int r)
    {
        // check if the indices lie in the range of string
        // and also if it is palindrome
        while (l >= 0 && r < s.Length && s[l] == s[r]) {
            // expand the boundary
            l--;
            r++;
        }
        // if it's length is greater than maxLength update
        // maxLength and res
        if (r - l - 1 >= maxLength) {
            res = s.Substring(l + 1, r - l - 1);
            maxLength = r - l - 1;
        }
        return;
    }
    // A function which takes a string prints the LPS and
    // returns the length of LPS
    public static int longestPalSubstr(string str)
    {
        res = "";
        maxLength = 1;
        // for every index in the string check palindromes
        // starting from that index
        for (int i = 0; i < str.Length; i++) {
            // check for odd length palindromes
            cSubUtil(str, i, i);
            // check for even length palindromes
            cSubUtil(str, i, i + 1);
        }
 
        Console.Write("Longest palindrome substring is: ");
        Console.WriteLine(res);
        return maxLength;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        string str = "forgeeksskeegfor";
        Console.WriteLine("Length is: "
                          + longestPalSubstr(str));
    }
}


Output

Longest palindrome substring is: geeksskeeg

Length is: 10

Please write comments if you find anything incorrect, or have more information about the topic discussed above.


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