1679. Max Number of K-Sum Pairs

Input: nums = [1,2,3,4], k = 5
Output: 2
Explanation: 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.

Input: nums = [3,1,3,4,3], k = 6
Output: 1
Explanation: Starting with nums = [3,1,3,4,3]:
- Remove the first two 3's, then nums = [1,4,3]
There are no more pairs that sum up to 6, hence a total of 1 operation.

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var maxOperations = function (nums, k) {
  const hashMap = new Map();
  let count = 0;
 
  for (let num of nums) {
    let complement = k - num;
    //if hashmap has complement of current num, simply decrement complement and increment count
    if (hashMap.has(complement) && hashMap.get(complement) > 0) {
      hashMap.set(complement, hashMap.get(complement) - 1);
      count++;
      //else add current num to hashmap, or increment if num already in hashmap
    } else {
      hashMap.has(num)
        ? hashMap.set(num, hashMap.get(num) + 1)
        : hashMap.set(num, 1);
    }
  }
 
  return count;
};

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var maxOperations = function (nums, k) {
  let addends = new Map();
  let operations = 0;
 
  for (let i = 0; i < nums.length; i++) {
    if (nums[i] >= k) continue;
 
    addends.has(nums[i])
      ? addends.set(nums[i], addends.get(nums[i]) + 1)
      : addends.set(nums[i], 1);
  }
 
  for (const [key, entry] of addends) {
    const complement = k - key;
    if (complement == key) operations += Math.floor(addends.get(key) / 2);
    else if (addends.has(complement)) {
      operations += Math.min(addends.get(complement), addends.get(key));
    }
 
    addends.delete(key);
    addends.delete(complement);
  }
 
  return operations;
};

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var maxOperations = function (nums, k) {
  let m = new Map(),
    ans = 0;
  for (let n of nums)
    if (n < k)
      if (m.get(k - n)) m.set(k - n, m.get(k - n) - 1), ans++;
      else m.set(n, (m.get(n) || 0) + 1);
  return ans;
};