Max Number of K-Sum Pairs
<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>In one operation, you can pick two numbers from the array whose sum equals <code>k</code> and remove them from the array.</p>

<p>Return <em>the maximum number of operations you can perform on the array</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> Starting with nums = [1,2,3,4]:
- Remove numbers 1 and 4, then nums = [2,3]
- Remove numbers 2 and 3, then nums = []
There are no more pairs that sum up to 5, hence a total of 2 operations.</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1,3,4,3], k = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> Starting with nums = [3,1,3,4,3]:
- Remove the first two 3&#39;s, then nums = [1,4,3]
There are no more pairs that sum up to 6, hence a total of 1 operation.</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>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Two Pointers; Sorting
impl Solution {
    pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
        let mut nums = nums.clone();
        nums.sort();
        let (mut l, mut r, mut ans) = (0, nums.len() - 1, 0);
        while l < r {
            match nums[l] + nums[r] {
                sum if sum == k => {
                    ans += 1;
                    l += 1;
                    r -= 1;
                }
                sum if sum > k => {
                    r -= 1;
                }
                _ => {
                    l += 1;
                }
            }
        }
        ans
    }
}
