<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>

class Solution {
public:
    int maxOperations(vector<int>& nums, int k) {
        // 1. remove num > k
        // 2. build num -> count hash table.
        // 3. iterate the hash table to count pairs.

        unordered_map<int, int> m; // num -> count
        int operations = 0;

        for (auto num : nums) {
            // the number which larger or equals than k can't be one of pair
            if (num < k) m[num]++;
        }
        
        for (auto [num, cnt] : m) {
            // No one can be current number's pair
            int remaining = k - num;
            if (m.find(remaining) == m.end()) continue;

            // pair count from the same number (ex: 3 + 3 = 6) or
            // different numbers (ex: 2 + 4 = 6)
            int pairCnt = num == remaining ? m[num] / 2 : min(m[num], m[remaining]);
            operations += pairCnt;
            m[num] -= pairCnt;
            m[remaining] -= pairCnt;
        }

        return operations;
    }
};
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