Count Deletions that Balance Index-Parity Sums
Company: GoDaddy
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Other
### Problem
Delete exactly one entry from `nums` and close the gap, giving the remaining entries new zero-based indices. Count the original indices whose deletion makes the new even-index sum equal to the new odd-index sum.
```cpp
class Solution {
public:
int waysToMakeFair(vector<int>& nums);
};
```
### Examples
**Example 1:**
```text
nums = [4,4,4]
Output: 3
Explanation: Each deletion leaves [4,4], whose parity sums match.
```
**Example 2:**
```text
nums = [2,2,5]
Output: 1
Explanation: Only deleting the final 5 leaves equal parity sums.
```
### Constraints
- `1 <= nums.length <= 10^5`.
- `1 <= nums[i] <= 10^4`.
### Clarifying Questions
Exactly one entry is removed. Entries to its right shift left one position, reversing their index parity; entries to its left keep their parity.
```hint Inspect the structure
How do positions to either side of a deletion acquire their new zero-based indices?
```
### Reasoning checkpoints
- Exclude the deleted value from both sides.
- Flip suffix parity before checking equality.
### Optional extensions
What answer follows for a single-element input, where the remaining array is empty?
Overview: Count Deletions that Balance Index-Parity Sums. Delete exactly one entry from nums and close the gap, giving the remaining entries new zero-based indices. Count the original indices whose deletion makes the new even-index sum equal to the new odd-index sum.