Check if Binary Array can be made palindrome after K bitwise XOR with 1

Last Updated : 14 Jun, 2022

Given a binary array arr[] of size N and an integer K, the task is to check whether the binary array can be turned into a palindrome after K number of operations where in one operation, any random index can be chosen and the value stored in the index will be replaced by its bitwise XOR(^) with1.

Examples:

Input: arr[] = { 1, 0, 0, 1, 1}, K = 0
Output: No
Explanation: As the array is not a palindrome, we must need some non-zero number of operations
to make it palindrome.

Input: arr[] = { 1, 0, 1, 1, 0, 0}, K = 1
Output: Yes
Explanation: Closest arrays which are palindrome : {0, 0, 1, 1, 0, 0} and {1, 0, 1, 1, 0, 1}, which is possible
using 1 such operations only.

Approach: The approach to solve the problem is based on the following idea:

The minimum number of operations required is the number of unequal elements equidistant from the mid of the array and to make them equal it takes 1 operation (from 1 to 0 or from 0 to 1).
There is need of 2 operations to change two equal elements and make them have the same value again using bitwise XOR.

To implement this idea, follow the steps shown below :

Initialize two pointers pointing at the two ends of the array.
Find the number of unequal array elements while traversing the array from its two ends.
If the number is greater than K, then it is impossible to reach the target with exactly K steps.
If the number is exactly equal to K, then it is possible to make the array palindrome.
Otherwise, if the number is lesser than K:
- If the array length is odd, then any number of positive K is fine as we can perform the operations at the mid index.
- If the array length is even, to keep it a palindrome, we must choose two indices and make the operations on those indices. So remaining K has to be even.

Here is the code for the above approach:

C++

   // C++ code for the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether the array
// can be turned into palindrome after K
// number of operations
bool check_possibility(int* arr, int K)
{
    // Store the length of the array in
    // arr_length variable
    int arr_length = sizeof(arr) / sizeof(
                                       arr[0]);
 
    // Initialize two pointers from left and
    // right ends of the array
    int i = 0;
    int j = arr_length - 1;
 
    // Keep iterating through the array from
    // two ends until the two pointers cross
    // each other to count the number
    // of the different array items.
    while (i < j) {
 
        // If the two elements are unequal,
        // decrease the value of K by one
        if (arr[i] != arr[j]) {
            K--;
        }
 
        // Move the left pointer towards the
        // right and right pointer towards the
        // left
        i++;
        j--;
    }
 
    // The unequal items are more than K or K
    // becomes less than zero, it is impossible
    // to make it a palindrome with D operations.
    if (K < 0) {
        return false;
    }
 
    // If K has a non-negative value, we make the
    // array a palindrome if it has an odd length
    // the remaining value of K is odd as we have
    // to choose two indices at a time to keep it
    // as a palindrome.
    else {
        if ((arr_length % 2) != 0
            || (K % 2) == 0) {
            return true;
        }
        else {
            return false;
        }
    }
}
 
// Driver code
int main()
{
    int arr[6] = { 1, 0, 1, 1, 0, 0 };
    int K = 1;
 
    // Function call
    if (check_possibility(arr, K) == true) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
    return 0;
}
 
  

Java

   // Java code for the approach
import java.io.*;
 
class GFG
{
 
  // Function to check whether the array
  // can be turned into palindrome after K
  // number of operations
  public static boolean check_possibility(int arr[],
                                          int K)
  {
 
    // Store the length of the array in
    // arr_length variable
    int arr_length = arr.length;
 
    // Initialize two pointers from left and
    // right ends of the array
    int i = 0;
    int j = arr_length - 1;
 
    // Keep iterating through the array from
    // two ends until the two pointers cross
    // each other to count the number
    // of the different array items.
    while (i < j) {
 
      // If the two elements are unequal,
      // decrease the value of K by one
      if (arr[i] != arr[j]) {
        K--;
      }
 
      // Move the left pointer towards the
      // right and right pointer towards the
      // left
      i++;
      j--;
    }
 
    // The unequal items are more than K or K
    // becomes less than zero, it is impossible
    // to make it a palindrome with D operations.
    if (K < 0) {
      return false;
    }
 
    // If K has a non-negative value, we make the
    // array a palindrome if it has an odd length
    // the remaining value of K is odd as we have
    // to choose two indices at a time to keep it
    // as a palindrome.
    else {
      if ((arr_length % 2) != 0 || (K % 2) == 0) {
        return true;
      }
      else {
        return false;
      }
    }
  }
  public static void main(String[] args)
  {
    int arr[] = { 1, 0, 1, 1, 0, 0 };
    int K = 1;
 
    // Function call
    if (check_possibility(arr, K) == true) {
      System.out.println("Yes");
    }
    else {
      System.out.println("No");
    }
  }
}
 
// This code is contributed by Rohit Pradhan
 
  

C#

   // C# code for the approach
using System;
class GFG {
 
  // Function to check whether the array
  // can be turned into palindrome after K
  // number of operations
  static bool check_possibility(int[] arr, int K)
  {
 
    // Store the length of the array in
    // arr_length variable
    int arr_length = arr.Length;
 
    // Initialize two pointers from left and
    // right ends of the array
    int i = 0;
    int j = arr_length - 1;
 
    // Keep iterating through the array from
    // two ends until the two pointers cross
    // each other to count the number
    // of the different array items.
    while (i < j) {
 
      // If the two elements are unequal,
      // decrease the value of K by one
      if (arr[i] != arr[j]) {
        K--;
      }
 
      // Move the left pointer towards the
      // right and right pointer towards the
      // left
      i++;
      j--;
    }
 
    // The unequal items are more than K or K
    // becomes less than zero, it is impossible
    // to make it a palindrome with D operations.
    if (K < 0) {
      return false;
    }
 
    // If K has a non-negative value, we make the
    // array a palindrome if it has an odd length
    // the remaining value of K is odd as we have
    // to choose two indices at a time to keep it
    // as a palindrome.
    else {
      if ((arr_length % 2) != 0 || (K % 2) == 0) {
        return true;
      }
      else {
        return false;
      }
    }
  }
  public static int Main()
  {
    int[] arr = new int[] { 1, 0, 1, 1, 0, 0 };
    int K = 1;
 
    // Function call
    if (check_possibility(arr, K) == true) {
      Console.WriteLine("Yes");
    }
    else {
      Console.WriteLine("No");
    }
    return 0;
  }
}
 
// This code is contributed by Taranpreet
 
  

Python3

   # Python code for the approach
 
def check_possibility(arr, K):
    
    # Store the length of the array in 
    # arr_length variable
    arr_length = len(arr)
     
    # Initialize two pointers from left and right 
    # ends of the array
    i = 0
    j = arr_length - 1
     
    # Keep iterating through the array from two 
    # ends until the two pointers cross each
    # other to count the number of the different 
    # array items.
    while(i < j):
       
        # If the two elements are unequal, 
        # decrease the value of K by one
        if(arr[i] != arr[j]):
            K -= 1
 
        # Move the left pointer towards the right
        # and right pointer towards the left
        i += 1
        j -= 1
     
    # The unequal items are more than K or
    # K becomes less than zero, it is impossible
    # to make it a palindrome with K operations.
    if(K < 0):
        return False
     
    # If K has a non-negative value, we make the 
    # array a palindrome if it has an odd length
    # the remaining value of K is odd as we have 
    # to choose two indices at a time 
    # to keep it as a palindrome
    else:
        if( (arr_length % 2)!= 0 or (K % 2)== 0):
            return True
     
        else:
            return False
 
# Driver code
if __name__ == '__main__':
    arr = [1, 0, 1, 1, 0, 0]
    K = 1
     
    if check_possibility(arr, K) == True :
        print("Yes")
    else:
        print("No")
 
  

Javascript

   // JavaScript code for the approach
 
// Function to check whether the array
// can be turned into palindrome after K
// number of operations
function check_possibility(arr, K)
{
 
    // Store the length of the array in
    // arr_length variable
    var arr_length = arr.length;
 
    // Initialize two pointers from left and
    // right ends of the array
    var i = 0;
    var j = arr_length - 1;
 
    // Keep iterating through the array from
    // two ends until the two pointers cross
    // each other to count the number
    // of the different array items.
    while (i < j) {
 
        // If the two elements are unequal,
        // decrease the value of K by one
        if (arr[i] != arr[j]) {
            K--;
        }
 
        // Move the left pointer towards the
        // right and right pointer towards the
        // left
        i++;
        j--;
    }
 
    // The unequal items are more than K or K
    // becomes less than zero, it is impossible
    // to make it a palindrome with D operations.
    if (K < 0) {
        return false;
    }
 
    // If K has a non-negative value, we make the
    // array a palindrome if it has an odd length
    // the remaining value of K is odd as we have
    // to choose two indices at a time to keep it
    // as a palindrome.
    else {
        if ((arr_length % 2) != 0 || (K % 2) == 0) {
            return true;
        }
        else {
            return false;
        }
    }
}
 
// Driver code
var arr = [ 1, 0, 1, 1, 0, 0 ];
var K = 1;
 
// Function call
if (check_possibility(arr, K) == true) {
    console.log("Yes");
}
else {
    console.log("No");
}
 
// This code is contributed by phasing17
 
  

Output

Yes

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

Check if given Array can be made a binary Array with K consecutive 1s

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Check if Binary Array can be made palindrome after K bitwise XOR with 1

C++

Java

C#

Python3

Javascript

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