Maximum Number of Ways to Partition an Array
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>. The number of ways to <strong>partition</strong> <code>nums</code> is the number of <code>pivot</code> indices that satisfy both conditions:</p>

<ul>
	<li><code>1 &lt;= pivot &lt; n</code></li>
	<li><code>nums[0] + nums[1] + ... + nums[pivot - 1] == nums[pivot] + nums[pivot + 1] + ... + nums[n - 1]</code></li>
</ul>

<p>You are also given an integer <code>k</code>. You can choose to change the value of <strong>one</strong> element of <code>nums</code> to <code>k</code>, or to leave the array <strong>unchanged</strong>.</p>

<p>Return <em>the <strong>maximum</strong> possible number of ways to <strong>partition</strong> </em><code>nums</code><em> to satisfy both conditions after changing <strong>at most</strong> one element</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [2,-1,2], k = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> One optimal approach is to change nums[0] to k. The array becomes [<strong><u>3</u></strong>,-1,2].
There is one way to partition the array:
- For pivot = 2, we have the partition [3,-1 | 2]: 3 + -1 == 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,0,0], k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The optimal approach is to leave the array unchanged.
There are two ways to partition the array:
- For pivot = 1, we have the partition [0 | 0,0]: 0 == 0 + 0.
- For pivot = 2, we have the partition [0,0 | 0]: 0 + 0 == 0.
</pre>

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

<pre>
<strong>Input:</strong> nums = [22,4,-25,-20,-15,15,-16,7,19,-10,0,-13,-14], k = -33
<strong>Output:</strong> 4
<strong>Explanation:</strong> One optimal approach is to change nums[2] to k. The array becomes [22,4,<u><strong>-33</strong></u>,-20,-15,15,-16,7,19,-10,0,-13,-14].
There are four ways to partition the array.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= k, nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Hash Table; Counting; Enumeration; Prefix Sum
func waysToPartition(nums []int, k int) (ans int) {
	n := len(nums)
	s := make([]int, n)
	s[0] = nums[0]
	right := map[int]int{}
	for i := range nums[:n-1] {
		right[s[i]]++
		s[i+1] = s[i] + nums[i+1]
	}
	if s[n-1]%2 == 0 {
		ans = right[s[n-1]/2]
	}
	left := map[int]int{}
	for i, x := range nums {
		d := k - x
		if (s[n-1]+d)%2 == 0 {
			t := left[(s[n-1]+d)/2] + right[(s[n-1]-d)/2]
			if ans < t {
				ans = t
			}
		}
		left[s[i]]++
		right[s[i]]--
	}
	return
}