1664. Ways to Make a Fair Array

Problem

You are given an integer array nums. You can choose exactly one index (0-indexed) and remove the element. Notice that the index of the elements may change after the removal.

For example, if nums = [6,1,7,4,1]:

• Choosing to remove index 1 results in nums = [6,7,4,1].

• Choosing to remove index 2 results in nums = [6,1,4,1].

• Choosing to remove index 4 results in nums = [6,1,7,4].

An array is fair if the sum of the odd-indexed values equals the sum of the even-indexed values.

Return the **number of indices that you could choose such that after the removal, nums ****is *fair*. **

  Example 1:

Input: nums = [2,1,6,4]
Output: 1
Explanation:
Remove index 0: [1,6,4] -> Even sum: 1 + 4 = 5. Odd sum: 6. Not fair.
Remove index 1: [2,6,4] -> Even sum: 2 + 4 = 6. Odd sum: 6. Fair.
Remove index 2: [2,1,4] -> Even sum: 2 + 4 = 6. Odd sum: 1. Not fair.
Remove index 3: [2,1,6] -> Even sum: 2 + 6 = 8. Odd sum: 1. Not fair.
There is 1 index that you can remove to make nums fair.

Example 2:

Input: nums = [1,1,1]
Output: 3
Explanation: You can remove any index and the remaining array is fair.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: You cannot make a fair array after removing any index.

  Constraints:

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^4

Solution (Java)

class Solution {
    public int waysToMakeFair(int[] nums) {
        int res = 0;
        int[] even = new int[nums.length];
        int[] odd = new int[nums.length];
        int oddSum = 0;
        int evenSum = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i % 2 == 0) {
                evenSum += nums[i];
            } else {
                oddSum += nums[i];
            }
            even[i] = evenSum;
            odd[i] = oddSum;
        }
        for (int i = 0; i < nums.length; i++) {
            if (i == 0) {
                evenSum = odd[nums.length - 1] - odd[0];
                oddSum = even[nums.length - 1] - even[0];
            } else {
                oddSum = odd[i - 1] + even[nums.length - 1] - even[i];
                evenSum = even[i - 1] + odd[nums.length - 1] - odd[i];
            }
            if (evenSum == oddSum) {
                res++;
            }
        }
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).