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# Test if each Subarray of even size has sum 0

Given an array A[ ] of measurement N, the duty is to test if the sum of each even-sized subarray is 0 or not.

Examples:

Enter: N = 4, A[] = {8, -8, 7, 9}
Output: NO
Clarification: Sum of subarray {7, 9} isn’t 0.

Enter: N = 2, A[] = {0, 0}
Output: YES
Clarification: The one doable even size subarray is {0, 0} and its sum is 0.

Naive Method: The essential option to clear up the issue is as follows:

Generate all doable even size subarrays and test if sum is 0 or not and return “YES” or “NO” accordingly.

Time Complexity: O(N2)
Auxiliary Area: O(1)

Environment friendly Method: To unravel the issue observe the beneath thought:

The thought is to test the entire array as soon as for all doable subarrays of size 2 as a result of all different evn sized subarrays of size higher than 2 could be made by combining subarrays of size 2. So if all subarrays of size 2 have sum 0, all different even sized subarrays can even have sum 0.

Comply with the steps talked about beneath to implement the concept:

• Begin iterating from i = 1 to N-1:
• Test if the sum of A[i] and A[i-1] is 0 or not.
• If it isn’t 0, return the reply as “NO” and no have to calculate additional.
• If the iteration is over and the situation is glad for all of the subarrays, return “YES” because the required reply.

Under is the implementation of the above method.

## C++

 ` `  `#embody ` `utilizing` `namespace` `std;` ` `  `string clear up(``int` `N, ``int` `A[])` `{` `    ``int` `ans = 1;` ` `  `    ` `    ` `    ``for` `(``int` `i = 1; i < N; i++) {` `        ``if` `(A[i] + A[i - 1] != 0) {` `            ``ans = 0;` `            ``break``;` `        ``}` `    ``}` `    ``if` `(ans)` `        ``return` `"YES"``;` `    ``return` `"NO"``;` `}` ` `  `int` `principal()` `{` `    ``int` `A[] = { 8, -8, 7, 9 };` `    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);` ` `  `    ` `    ``cout << clear up(N, A);` `    ``return` `0;` `}`

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