Minimum Operations to Reduce X to Zero
<p>You are given an integer array <code>nums</code> and an integer <code>x</code>. In one operation, you can either remove the leftmost or the rightmost element from the array <code>nums</code> and subtract its value from <code>x</code>. Note that this <strong>modifies</strong> the array for future operations.</p>

<p>Return <em>the <strong>minimum number</strong> of operations to reduce </em><code>x</code> <em>to <strong>exactly</strong></em> <code>0</code> <em>if it is possible</em><em>, otherwise, return </em><code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,4,2,3], x = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> The optimal solution is to remove the last two elements to reduce x to zero.
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,6,7,8,9], x = 4
<strong>Output:</strong> -1
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,20,1,1,3], x = 10
<strong>Output:</strong> 5
<strong>Explanation:</strong> The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
</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>
	<li><code>1 &lt;= x &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Binary Search; Prefix Sum; Sliding Window
use std::collections::HashMap;

impl Solution {
    pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
        let s = nums.iter().sum::<i32>() - x;
        let mut vis: HashMap<i32, i32> = HashMap::new();
        vis.insert(0, -1);
        let mut mx = -1;
        let mut t = 0;
        for (i, v) in nums.iter().enumerate() {
            t += v;
            if !vis.contains_key(&t) {
                vis.insert(t, i as i32);
            }
            if let Some(&j) = vis.get(&(t - s)) {
                mx = mx.max((i as i32) - j);
            }
        }
        if mx == -1 {
            -1
        } else {
            (nums.len() as i32) - mx
        }
    }
}
