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
class Solution {
    public int waysToPartition(int[] nums, int k) {
        int n = nums.length;
        int[] s = new int[n];
        s[0] = nums[0];
        Map<Integer, Integer> right = new HashMap<>();
        for (int i = 0; i < n - 1; ++i) {
            right.merge(s[i], 1, Integer::sum);
            s[i + 1] = s[i] + nums[i + 1];
        }
        int ans = 0;
        if (s[n - 1] % 2 == 0) {
            ans = right.getOrDefault(s[n - 1] / 2, 0);
        }
        Map<Integer, Integer> left = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            int d = k - nums[i];
            if ((s[n - 1] + d) % 2 == 0) {
                int t = left.getOrDefault((s[n - 1] + d) / 2, 0)
                    + right.getOrDefault((s[n - 1] - d) / 2, 0);
                ans = Math.max(ans, t);
            }
            left.merge(s[i], 1, Integer::sum);
            right.merge(s[i], -1, Integer::sum);
        }
        return ans;
    }
}