<p>You are given an integer array&nbsp;<code>nums</code>. You can choose <strong>exactly one</strong> index (<strong>0-indexed</strong>) and remove the element. Notice that the index of the elements may change after the removal.</p>

<p>For example, if <code>nums = [6,1,7,4,1]</code>:</p>

<ul>
	<li>Choosing to remove index <code>1</code> results in <code>nums = [6,7,4,1]</code>.</li>
	<li>Choosing to remove index <code>2</code> results in <code>nums = [6,1,4,1]</code>.</li>
	<li>Choosing to remove index <code>4</code> results in <code>nums = [6,1,7,4]</code>.</li>
</ul>

<p>An array is <strong>fair</strong> if the sum of the odd-indexed values equals the sum of the even-indexed values.</p>

<p>Return the <em><strong>number</strong> of indices that you could choose such that after the removal, </em><code>nums</code><em> </em><em>is <strong>fair</strong>. </em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong>&nbsp;You can remove any index and the remaining array is fair.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 0
<strong>Explanation:</strong>&nbsp;You cannot make a fair array after removing any index.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

#if 0


[6,1,7,4,1]

PS = [6,1,13,5,14]

if we remove a value at index i which is even:
  the sum of even values is equal to PS[i] - val[i] + odd_sum - PS[i-1]
  the sum of odd values is equal to PS[i-1] + even_sum - PS[i]


#endif

class Solution {
public:
  int waysToMakeFair(vector<int>& nums) {
    vector<int> prefix_sum_interl;
    prefix_sum_interl.reserve(nums.size());
    int odd_sum{0};
    int even_sum{0};

    for (int i{0}; i < nums.size(); ++i) {
      if (i % 2) {
        odd_sum += nums[i];
        prefix_sum_interl.push_back(odd_sum);
      } else {
        even_sum += nums[i];
        prefix_sum_interl.push_back(even_sum);
      }
    }

    int num_ways{0};
    for (int i{0}; i < nums.size(); ++i) {
      if (i % 2) {
        int new_odd_sum{ prefix_sum_interl[i] - nums[i] + even_sum - prefix_sum_interl[i-1] };
        int new_even_sum{ prefix_sum_interl[i-1] + odd_sum - prefix_sum_interl[i] };        

        if (new_even_sum == new_odd_sum) {
          ++num_ways;
        }
      } else {
        int new_even_sum{ prefix_sum_interl[i] - nums[i] + odd_sum - (i == 0 ? 0 : prefix_sum_interl[i-1]) };
        int new_odd_sum{ (i == 0 ? 0 : prefix_sum_interl[i-1]) + even_sum - prefix_sum_interl[i] };

        if (new_even_sum == new_odd_sum) {
          ++num_ways;
        }
      }
    }

    return num_ways;
  }
};
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