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Check if given string can be made Palindrome by removing only single type of character

  • Difficulty Level : Easy
  • Last Updated : 23 Nov, 2021

Given a string S, the task is to whether a string can be made palindrome after removing the occurrences of the same character, any number of times

Examples:

 Input: S = “abczdzacb”
 Output: Yes
 Explanation: Remove first and second occurrence of character ‘a’, string S becomes “bczdzcb”, which is a palindrome . 
 
Input: S = “madem”
Output: No

 

Approach: The task can be solved by iterating over each unique character in the given string, and removing its occurrences wherever there is a mismatch, if a valid palindrome is found, after removing occurrences of the same character any number of times, return “Yes” else return “No“.
Follow the below steps to solve the problem:

  • Start iterating over each unique character of the string, whose occurrences are to be deleted
  • Use the two-pointer technique, to check for a mismatch, Place l at the start of the string and r at the end of the string
  • If S[l] == S[r], increment l, and decrement r.
  • If S[l]!= S[r], check if S[l[ == char, do l++, else if S[r] == char, do r–
  • If none of the conditions hold, means the given can’t be converted into a palindrome

 Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a palindrome is
// possible or not
string isPossible(string S)
{
    // Stores the length of string
    int n = (int)S.length();
 
    // Stores the unique characters in
    // the string
    set<char> st;
 
    for (int i = 0; i < n; i++) {
        st.insert(S[i]);
    }
 
    // Check if valid palindrome is
    // possible or not
    bool check = false;
 
    // Iterating over unique characters
    // of the string
    for (auto ele : st) {
 
        // Pointers to check the condition
        int low = 0, high = n - 1;
        bool flag = true;
 
        // Iterating over the string
        for (int i = 0; i < n; i++) {
            if (S[low] == S[high]) {
 
                // Updating low and high
                low++;
                high--;
            }
 
            else {
                if (S[low] == ele) {
 
                    // Updating low
                    low++;
                }
                else if (S[high] == ele) {
 
                    // Updating high
                    high--;
                }
                else {
 
                    // It is impossible
                    // to make palindrome
                    // by removing
                    // occurrences of char
                    flag = false;
                    break;
                }
            }
        }
 
        // If palindrome is formed
        // break the loop
        if (flag == true) {
            check = true;
            break;
        }
    }
 
    if (check)
        return "Yes";
    else
        return "No";
}
 
// Driver Code
int main()
{
 
    string S = "abczdzacb";
    cout << isPossible(S);
    return 0;
}
Java



// Java code for the above approach
import java.util.*;
 
class GFG
{
   
  // Function to check if a palindrome is
// possible or not
static String isPossible(String S)
{
   
    // Stores the length of string
    int n = S.length();
 
    // Stores the unique characters in
    // the string
    Set<Character> st = new HashSet<Character>();
     
    for (int i = 0; i < n; i++) {
        st.add(S.charAt(i));
    }
     
    // Check if valid palindrome is
    // possible or not
    boolean check = false;
 
    // Iterating over unique characters
    // of the string
    for (Character ele : st) {
 
        // Pointers to check the condition
        int low = 0, high = n - 1;
        boolean flag = true;
 
        // Iterating over the string
        for (int i = 0; i < n; i++) {
            if (S.charAt(low) == S.charAt(high)) {
 
                // Updating low and high
                low++;
                high--;
            }
 
            else {
                if (S.charAt(low) == ele) {
 
                    // Updating low
                    low++;
                }
                else if (S.charAt(high)== ele) {
 
                    // Updating high
                    high--;
                }
                else {
 
                    // It is impossible
                    // to make palindrome
                    // by removing
                    // occurrences of char
                    flag = false;
                    break;
                }
            }
        }
 
        // If palindrome is formed
        // break the loop
        if (flag == true) {
            check = true;
            break;
        }
    }
 
    if (check)
        return "Yes";
    else
        return "No";
}
 
// Driver Code
    public static void main (String[] args) {
      String S = "abczdzacb";
     
        System.out.println(isPossible(S));
    }
}
 
// This code is contributed by Potta Lokesh
Python3



# python program for the above approach
 
# Function to check if a palindrome is
# possible or not
def isPossible(S):
 
    # Stores the length of string
    n = len(S)
 
    # Stores the unique characters in
    # the string
    st = set()
 
    for i in range(0, n):
        st.add(S[i])
 
    # Check if valid palindrome is
    # possible or not
    check = False
 
    # Iterating over unique characters
    # of the string
    for ele in st:
 
        # Pointers to check the condition
        low = 0
        high = n - 1
        flag = True
 
        # Iterating over the string
        for i in range(0, n):
            if (S[low] == S[high]):
 
                # Updating low and high
                low += 1
                high -= 1
 
            else:
                if (S[low] == ele):
 
                    # Updating low
                    low += 1
 
                elif (S[high] == ele):
 
                    # Updating high
                    high -= 1
 
                else:
 
                    # It is impossible
                    # to make palindrome
                    # by removing
                    # occurrences of char
                    flag = False
                    break
 
                # If palindrome is formed
                # break the loop
        if (flag == True):
            check = True
            break
 
    if (check):
        return "Yes"
 
    else:
        return "No"
 
# Driver Code
if __name__ == "__main__":
 
    S = "abczdzacb"
    print(isPossible(S))
 
    # This code is contributed by rakeshsahni
C#



// C# code for the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
   
  // Function to check if a palindrome is
// possible or not
static String isPossible(String S)
{
   
    // Stores the length of string
    int n = S.Length;
 
    // Stores the unique characters in
    // the string
    HashSet<char> st = new HashSet<char>();
     
    for (int i = 0; i < n; i++) {
        st.Add(S[i]);
    }
     
    // Check if valid palindrome is
    // possible or not
    bool check = false;
 
    // Iterating over unique characters
    // of the string
    foreach (char ele in st) {
 
        // Pointers to check the condition
        int low = 0, high = n - 1;
        bool flag = true;
 
        // Iterating over the string
        for (int i = 0; i < n; i++) {
            if (S[low] == S[high]) {
 
                // Updating low and high
                low++;
                high--;
            }
 
            else {
                if (S[low] == ele) {  
 
                    // Updating low
                    low++;
                }
                else if (S[high]== ele) {
 
                    // Updating high
                    high--;
                }
                else {
 
                    // It is impossible
                    // to make palindrome
                    // by removing
                    // occurrences of char
                    flag = false;
                    break;
                }
            }
        }
 
        // If palindrome is formed
        // break the loop
        if (flag == true) {
            check = true;
            break;
        }
    }
 
    if (check)
        return "Yes";
    else
        return "No";
}
 
// Driver Code
    public static void Main(String[] args) {
      String S = "abczdzacb";
     
        Console.WriteLine(isPossible(S));
    }
}
 
// This code is contributed by shikhasingrajput
Javascript



<script>
// Javascript program for the above approach
 
// Function to check if a palindrome is
// possible or not
function isPossible(S)
{
 
  // Stores the length of string
  let n = S.length;
 
  // Stores the unique characters in
  // the string
  let st = new Set();
 
  for (let i = 0; i < n; i++) {
    st.add(S[i]);
  }
 
  // Check if valid palindrome is
  // possible or not
  let check = false;
 
  // Iterating over unique characters
  // of the string
  for (ele of st) {
 
    // Pointers to check the condition
    let low = 0, high = n - 1;
    let flag = true;
 
    // Iterating over the string
    for (let i = 0; i < n; i++) {
      if (S[low] == S[high]) {
 
        // Updating low and high
        low++;
        high--;
      }
 
      else {
        if (S[low] == ele) {
 
          // Updating low
          low++;
        }
        else if (S[high] == ele) {
 
          // Updating high
          high--;
        }
        else {
 
          // It is impossible
          // to make palindrome
          // by removing
          // occurrences of char
          flag = false;
          break;
        }
      }
    }
 
    // If palindrome is formed
    // break the loop
    if (flag == true) {
      check = true;
      break;
    }
  }
 
  if (check)
    return "Yes";
  else
    return "No";
}
 
// Driver Code
let S = "abczdzacb";
document.write(isPossible(S));
 
// This code is contributed by saurabh_jaiswal.
</script>
 
 
Output: 
Yes
 
 
Time Complexity: O(n*26)
Auxiliary Space:  O(n) 
 

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Article Contributed By :
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subhankarjadab
@subhankarjadab
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