Skip to content
Related Articles

Related Articles

Improve Article

Longest palindromic string possible after removal of a substring

  • Last Updated : 13 Aug, 2021
Geek Week

Given a string str, the task is to find the longest palindromic string that can be obtained from it after removing a substring.

Examples: 

Input: str = “abcdefghiedcba” 
Output: “abcdeiedcba” 
Explanation: Removal of substring “fgh” leaves the remaining string palindromic

Input: str = “abba” 
Output: “abba” 
Explanation: Removal of substring “” as the given string is already palindromic. 

Approach: 



  • Find the longest possible pair of substrings A and B from both ends of the given string which are reverse of each other.
  • Remove them from the original string.
  • Find the longest palindromic substrings from both ends of the remaining string using KMP and consider the substring which is longer.
  • Add the strings A and B to beginning and end of this palindromic substring respectively to get the desired output.

Below code 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 longest palindrome
// from the start of the string using KMP match
string findPalindrome(string C)
{
    string S = C;
    reverse(S.begin(), S.end());
    // Append S(reverse of C)  to C
    C = C + "&" + S;
    int n = C.length();
    int longestPalindrome[n];
    longestPalindrome[0] = 0;
    int len = 0;
    int i = 1;
    // Use KMP algorithm
    while (i < n) {
        if (C[i] == C[len]) {
            len++;
            longestPalindrome[i] = len;
            i++;
        }
        else {
            if (len != 0) {
                len = longestPalindrome[len - 1];
            }
            else {
                longestPalindrome[i] = 0;
                i++;
            }
        }
    }
    string ans = C.substr(0, longestPalindrome[n - 1]);
    return ans;
}
 
// Function to return longest palindromic
// string possible from the given string
// after removal of any substring
string findAns(string s)
{
    // Initialize three strings A, B AND F
    string A = "";
    string B = "";
    string F = "";
 
    int i = 0;
    int j = s.length() - 1;
    int len = s.length();
     
    // Loop to find longest substrings
    // from both ends which are
    // reverse of each other
    while (i < j && s[i] == s[j]) {
        i = i + 1;
        j = j - 1;
    }
     
    if (i > 0)
    {
        A = s.substr(0, i);
        B = s.substr(len - i, i);
    }
  
    // Proceed to third step of our approach
    if (len > 2 * i)
    {
        // Remove the substrings A and B
        string C = s.substr(i, s.length() - 2 * i);
        // Find the longest palindromic
        // substring from beginning of C
        string D = findPalindrome(C);
         
        // Find the longest palindromic
        // substring from end of C
        reverse(C.begin(), C.end());
        string E = findPalindrome(C);
         
        // Store the maximum of D and E in F
        if (D.length() > E.length()) {
            F = D;
        }
        else {
            F = E;
        }
    }
     
    // Find the final answer
    string answer = A + F + B;
     
    return answer;
}
// Driver Code
int main()
{
    string str = "abcdefghiedcba";
    cout << findAns(str) << endl;
  
}

Java




// Java Implementation of the
// above approach
import java.util.*;
 
class GFG{
     
// Function to find the longest palindrome
// from the start of the String using KMP match
static String findPalindrome(String C)
{
    String S = C;
    S = reverse(S);
     
    // Append S(reverse of C) to C
    C = C + "&" + S;
    int n = C.length();
    int []longestPalindrome = new int[n];
    longestPalindrome[0] = 0;
    int len = 0;
    int i = 1;
     
    // Use KMP algorithm
    while (i < n) {
        if (C.charAt(i) == C.charAt(len)) {
            len++;
            longestPalindrome[i] = len;
            i++;
        }
        else {
            if (len != 0) {
                len = longestPalindrome[len - 1];
            }
            else {
                longestPalindrome[i] = 0;
                i++;
            }
        }
    }
    String ans = C.substring(0, longestPalindrome[n - 1]);
    return ans;
}
 
// Function to return longest palindromic
// String possible from the given String
// after removal of any subString
static String findAns(String s)
{
    // Initialize three Strings A, B AND F
    String A = "";
    String B = "";
    String F = "";
 
    int i = 0;
    int j = s.length() - 1;
    int len = s.length();
     
    // Loop to find longest subStrings
    // from both ends which are
    // reverse of each other
    while (i < j && s.charAt(i) == s.charAt(j)) {
        i = i + 1;
        j = j - 1;
    }
     
    if (i > 0)
    {
        A = s.substring(0, i);
        B = s.substring(len - i, len);
    }
 
    // Proceed to third step of our approach
    if (len > 2 * i)
    {
        // Remove the subStrings A and B
        String C = s.substring(i, (s.length() - 2 * i) + i);
         
        // Find the longest palindromic
        // subString from beginning of C
        String D = findPalindrome(C);
         
        // Find the longest palindromic
        // subString from end of C
        C = reverse(C);
        String E = findPalindrome(C);
         
        // Store the maximum of D and E in F
        if (D.length() > E.length()) {
            F = D;
        }
        else {
            F = E;
        }
    }
     
    // Find the final answer
    String answer = A + F + B;
     
    return answer;
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "abcdefghiedcba";
    System.out.print(findAns(str) +"\n");
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the
# above approach
 
# Function to find the longest
# palindrome from the start of
# the string using KMP match
def findPalindrome(C):
     
    S = C[::-1]
     
    # Append S(reverse of C)  to C
    C = C[:] + '&' + S
     
    n = len(C)
    longestPalindrome = [0 for i in range(n)]
    longestPalindrome[0] = 0
     
    ll = 0
    i = 1
     
    # Use KMP algorithm
    while (i < n):
        if (C[i] == C[ll]):
            ll += 1
            longestPalindrome[i] = ll
            i += 1
         
        else:
             
            if (ll != 0):
                ll = longestPalindrome[ll - 1]
            else:
                longestPalindrome[i] = 0
                i += 1
             
    ans = C[0:longestPalindrome[n - 1]]
     
    return ans
 
# Function to return longest palindromic
# string possible from the given string
# after removal of any substring
def findAns(s):
 
    # Initialize three strings
    # A, B AND F
    A = ""
    B = ""
    F = ""
   
    i = 0
    j = len(s) - 1
    ll = len(s)
       
    # Loop to find longest substrings
    # from both ends which are 
    # reverse of each other
    while (i < j and s[i] == s[j]):
        i = i + 1
        j = j - 1
     
    if (i > 0):
        A = s[0 : i]
        B = s[ll - i : ll]
      
    # Proceed to third step of our approach
    if (ll > 2 * i): 
     
        # Remove the substrings A and B
        C = s[i : i + (len(s) - 2 * i)]
         
        # Find the longest palindromic
        # substring from beginning of C
        D = findPalindrome(C)
           
        # Find the longest palindromic
        # substring from end of C
        C = C[::-1]
         
        E = findPalindrome(C)
           
        # Store the maximum of D and E in F
        if (len(D) > len(E)):
            F = D
        else:
            F = E
     
    # Find the final answer
    answer = A + F + B
       
    return answer
 
# Driver code
if __name__=="__main__":
     
    str = "abcdefghiedcba"
     
    print(findAns(str))
 
# This code is contributed by rutvik_56

C#




// C# Implementation of the
// above approach
using System;
 
class GFG{
      
// Function to find the longest palindrome
// from the start of the String using KMP match
static String findPalindrome(String C)
{
    String S = C;
    S = reverse(S);
      
    // Append S(reverse of C) to C
    C = C + "&" + S;
    int n = C.Length;
    int []longestPalindrome = new int[n];
    longestPalindrome[0] = 0;
    int len = 0;
    int i = 1;
      
    // Use KMP algorithm
    while (i < n) {
        if (C[i] == C[len]) {
            len++;
            longestPalindrome[i] = len;
            i++;
        }
        else {
            if (len != 0) {
                len = longestPalindrome[len - 1];
            }
            else {
                longestPalindrome[i] = 0;
                i++;
            }
        }
    }
    String ans = C.Substring(0, longestPalindrome[n - 1]);
    return ans;
}
  
// Function to return longest palindromic
// String possible from the given String
// after removal of any subString
static String findAns(String s)
{
    // Initialize three Strings A, B AND F
    String A = "";
    String B = "";
    String F = "";
  
    int i = 0;
    int j = s.Length - 1;
    int len = s.Length;
      
    // Loop to find longest subStrings
    // from both ends which are
    // reverse of each other
    while (i < j && s[i] == s[j]) {
        i = i + 1;
        j = j - 1;
    }
      
    if (i > 0)
    {
        A = s.Substring(0, i);
        B = s.Substring(len - i, i);
    }
  
    // Proceed to third step of our approach
    if (len > 2 * i)
    {
        // Remove the subStrings A and B
        String C = s.Substring(i, (s.Length - 2 * i));
          
        // Find the longest palindromic
        // subString from beginning of C
        String D = findPalindrome(C);
          
        // Find the longest palindromic
        // subString from end of C
        C = reverse(C);
        String E = findPalindrome(C);
          
        // Store the maximum of D and E in F
        if (D.Length > E.Length) {
            F = D;
        }
        else {
            F = E;
        }
    }
      
    // Find the readonly answer
    String answer = A + F + B;
      
    return answer;
}
static String reverse(String input) {
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "abcdefghiedcba";
    Console.Write(findAns(str) +"\n");
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
      // JavaScript Implementation of the
      // above approach
      // Function to find the longest palindrome
      // from the start of the String using KMP match
      function findPalindrome(C) {
        var S = C;
        S = reverse(S);
 
        // Append S(reverse of C) to C
        C = C + "&" + S;
        var n = C.length;
        var longestPalindrome = new Array(n).fill(0);
        longestPalindrome[0] = 0;
        var len = 0;
        var i = 1;
 
        // Use KMP algorithm
        while (i < n) {
          if (C[i] === C[len]) {
            len++;
            longestPalindrome[i] = len;
            i++;
          } else {
            if (len !== 0) {
              len = longestPalindrome[len - 1];
            } else {
              longestPalindrome[i] = 0;
              i++;
            }
          }
        }
        var ans = C.substring(0, longestPalindrome[n - 1]);
        return ans;
      }
 
      // Function to return longest palindromic
      // String possible from the given String
      // after removal of any subString
      function findAns(s) {
        // Initialize three Strings A, B AND F
        var A = "";
        var B = "";
        var F = "";
 
        var i = 0;
        var j = s.length - 1;
        var len = s.length;
 
        // Loop to find longest subStrings
        // from both ends which are
        // reverse of each other
        while (i < j && s[i] === s[j]) {
          i = i + 1;
          j = j - 1;
        }
 
        if (i > 0) {
          A = s.substring(0, i);
          B = s.substring(len - i, len);
        }
 
        // Proceed to third step of our approach
        if (len > 2 * i) {
          // Remove the subStrings A and B
          var C = s.substring(i, i + (s.length - 2 * i));
 
          // Find the longest palindromic
          // subString from beginning of C
          var D = findPalindrome(C);
 
          // Find the longest palindromic
          // subString from end of C
          C = reverse(C);
          var E = findPalindrome(C);
 
          // Store the maximum of D and E in F
          if (D.length > E.length) {
            F = D;
          } else {
            F = E;
          }
        }
 
        // Find the readonly answer
        var answer = A + F + B;
 
        return answer;
      }
 
      function reverse(input) {
        var a = input.split("");
        var r = a.length - 1;
        for (var l = 0; l < r; l++, r--) {
          var temp = a[l];
          a[l] = a[r];
          a[r] = temp;
        }
        return a.join("");
      }
 
      // Driver Code
      var str = "abcdefghiedcba";
      document.write(findAns(str));
       
      // This code is contributed by rdtank.
    </script>
Output: 
abcdeiedcba

 

Time complexity: O(N)
Auxiliary Space: O(N) 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :