# Examine if Binary Array might be made Palindrome by flipping two bits at a time

Given a binary array A[] of measurement N, the duty is to examine whether or not the array might be transformed right into a palindrome by flipping two bits in every operation.

Notice: The operation might be carried out any variety of occasions.

Examples:

Enter: A[] = {1, 0, 1, 0, 1, 1}
Output: Sure
Clarification: We are able to carry out the next operation:
Choose i = 2 and j = 4. Then {1, 0, 1, 0, 1, 1}→{1, 0, 0, 0, 0, 1} which is a palindrome.

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

Strategy: The issue might be solved based mostly on the next statement:

Depend variety of 0s and 1s.

• If the worth of depend of 0s and 1s are odd then no palindrome might be made by performing operation on A[].
• In any other case, array A[] can transformed right into a palindrome after performing operation on array.

Comply with the under steps to unravel the issue:

• Set zeros = 0 and ones = 0 to retailer the depend of variety of 0s and 1s within the array.
• After that iterate a loop from 0 to N-1 corresponding to:
• If A[i] = 0 increment the worth of zeros in any other case increment the worth of ones.
• If the worth of zeros and ones is odd then print “No” i.e. it isn’t potential to transform A to a palindrome by making use of the above operation any variety of occasions.
• In any other case, print “Sure”.

Beneath is the implementation of the above method.

## Java

 ` `  `import` `java.io.*;` `import` `java.util.*;` ` `  `public` `class` `GFG {` ` `  `    ` `    ` `    ``public` `static` `String examine(``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` `"Sure"``;` `    ``}` ` `  `    ` `    ``public` `static` `void` `fundamental(String[] args)` `    ``{` `        ``int``[] A = { ``1``, ``0``, ``1``, ``0``, ``1``, ``1` `};` `        ``int` `N = A.size;` ` `  `        ` `        ``System.out.println(examine(A, N));` `    ``}` `}`

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