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

Return the maximum number of operations you can perform on the array.

Example 1:

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.

Example 2:

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.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= k <= 109

Solution 1: Frequency Map

For each x, check freq[x] and freq[k – x]. Note: there is a special case when x + x == k.

Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
  int maxOperations(vector<int>& nums, int k) {
    unordered_map<int, int> m;
    int ans = 0;
    for (int x : nums) ++m[x];
    for (int x : nums) {      
      if (m[x] < 1 || m[k - x] < 1 + (x + x == k)) continue;
      --m[x];
      --m[k - x];
      ++ans;      
    }
    return ans;
  }
};

class Solution:
  def maxOperations(self, nums: List[int], k: int) -> int:
    m = defaultdict(int)
    ans = 0
    for x in nums: m[x] += 1
    for x in nums:
      if m[x] < 1 or m[k - x] < 1 + (x + x == k): continue
      m[x] -= 1
      m[k - x] -= 1
      ans += 1
    return ans

Solution 2: Two Pointers

Sort the number, start from i = 0, j = n – 1, compare s = nums[i] + nums[j] with k and move i, j accordingly.

Time complexity: O(nlogn)
Space complexity: O(1)

class Solution {
public:
  int maxOperations(vector<int>& nums, int k) {
    sort(begin(nums), end(nums));
    int i = 0, j = nums.size() - 1;
    int ans = 0;
    while (i < j) {
      const int s = nums[i] + nums[j];
      if (s == k) {
        ++ans;
        ++i; --j;
      } else if (s < k) {
        ++i;
      } else {
        --j;
      }
    }
    return ans;
  }
};

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