Check if Binary Array can be made Palindrome by flipping two bits at a time

Given a binary array A[] of size N, the task is to check whether the array can be converted into a palindrome by flipping two bits in each operation.

Note: The operation can be performed any number of times.

Examples:

Input: A[] = {1, 0, 1, 0, 1, 1}
Output: Yes

Explanation: We can perform the following operation:
Select i = 2 and j = 4. Then {1, 0, 1, 0, 1, 1}→{1, 0, 0, 0, 0, 1} which is a palindrome.

Input: A[] = {0, 1}
Output: No

Approach: The problem can be solved based on the following observation: 

Count number of 0s and 1s. 

• If the value of count of 0s and 1s are odd then no palindrome can be made by performing operation on A[].
• Otherwise, array A[] can converted into a palindrome after performing operation on array.

Follow the below steps to solve the problem:

• Set zeros = 0 and ones = 0 to store the count of number of 0s and 1s in the array.
• After that iterate a loop from 0 to N-1 such as:
  • If A[i] = 0 increment the value of zeros otherwise increment the value of ones.
• If the value of zeros and ones is odd then print “No” i.e. it is not possible to convert A to a palindrome by applying the above operation any number of times.
• Otherwise, print “Yes”.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find check whether
// array can converted into palindrome
string check(int arr[], int n)
{
    int one = 0, zero = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == '0') {
            zero++;
        }
        else {
            one++;
        }
    }
    if (one % 2 != 0 && zero % 2 != 0)
        return "No";
    return "Yes";
}

// Driver Code
int main()
{
    int A[] = { 1, 0, 1, 0, 1, 1 };
    int N = sizeof(A) / sizeof(A[0]);

    // Function call
    cout << check(A, N);

    return 0;
}

// This code is contributed by aarohirai2616.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

public class GFG {

    // Function to find check whether
    // array can converted into palindrome
    public static String check(int arr[], int n)
    {
        int one = 0, zero = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] == '0') {
                zero++;
            }
            else {
                one++;
            }
        }
        if (one % 2 != 0 && zero % 2 != 0)
            return "No";
        return "Yes";
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] A = { 1, 0, 1, 0, 1, 1 };
        int N = A.length;

        // Function Call
        System.out.println(check(A, N));
    }
}

Python3

# Python code to implement the approach

# Function to find check whether
# array can converted into palindrome
def check(arr, n):
    one = 0
    zero = 0
    for i in range(0, n):
            if (arr[i] == '0'):
                zero = zero + 1

            else:
                one = one + 1

    if(one % 2 != 0 & zero % 2 != 0):
            return "No"
    return "Yes"
  
# Driver Code
if __name__ == '__main__':
    A = [1, 0, 1, 0, 1, 1 ]
    N = len(A)

    # Function call
    print(check(A,N))

    # This code is contributed by aarohirai2616.

C#

using System;
class GFG {
  static string check(int[] arr, int n)
  {
    int one = 0, zero = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] == '0') {
        zero++;
      }
      else {
        one++;
      }
    }
    if (one % 2 != 0 && zero % 2 != 0)
      return "No";
    return "Yes";
  }
  static void Main()
  {
    int[] A = { 1, 0, 1, 0, 1, 1 };
    int N = 6;

    // Function Call
    Console.Write(check(A, N));
  }
}

// This code is contributed by garg28harsh.

Javascript

        // JavaScript code to implement the approach

        // Function to find check whether
        // array can converted into palindrome
        const check = (arr, n) => {
            let one = 0, zero = 0;
            for (let i = 0; i < n; i++) {
                if (arr[i] == '0') {
                    zero++;
                }
                else {
                    one++;
                }
            }
            if (one % 2 != 0 && zero % 2 != 0)
                return "No";
            return "Yes";
        }

        // Driver Code
        let A = [1, 0, 1, 0, 1, 1];
        let N = A.length;

        // Function call
        console.log(check(A, N));

        // This code is contributed by rakeshsahni

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

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