Longest palindromic string possible after removal of a substring


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++

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// 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;
   
}

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Java

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// 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

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C#

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// 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

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Output:

abcdeiedcba

Time complexity: O(N)

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Improved By : princiraj1992, Rajput-Ji