You are given an integer array nums and an integer k.

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 <= 10^5
1 <= nums[i] <= 10^9
1 <= k <= 10^9

Approach 1: Hash Map

We can use hash map to store the occurrences for each number. Just like Two Sum problem, we find the complement $k - x$ . Then, we will be facing two cases.

The first case is when $x$ is same as $k - x$ . In this case, we can only take $m[x] / 2$ at most, then we update both value with the remainder.

Otherwise, we can only take the minimal value of $m[x]$ and $m[k - x]$ , then update the hash map values.

class Solution {
public:
    int maxOperations(vector<int>& nums, int k) {
        unordered_map<int, int> m;
        int ans = 0;
        for (auto &x : nums) m[x]++;
        for (auto &x : nums) {
            if (m[k - x] > 0) {
                if (x == k - x) {
                    ans += m[x] / 2;
                    m[x] = m[x] % 2;
                    m[k - x] = m[x] % 2;
                } else {
                    int ops = min(m[x], m[k - x]);
                    ans += ops;
                    m[x] -= ops;
                    m[k - x] -= ops;
                }
            }
        }
        return ans;
    }
};

class Solution {
    public int maxOperations(int[] nums, int k) {
        int count = 0;
        Map<Integer, Integer> table = new HashMap<>();
        for (int x : nums) {
            int reminder = k - x;
            if (table.containsKey(reminder)) {
                count += 1;
                if (table.get(reminder) == 1) {
                    table.remove(reminder);
                } else {
                    table.merge(reminder, -1, Integer::sum);
                }
            } else {
                table.merge(x, 1, Integer::sum);
            }
        }
        return count;
    }
}

We can sort the input and use two pointers to track elements from both side. If their sum is equal to $k$ , then we increase $ans$ by 1. If $nums[i] + nums[j] > k$ , it means we should reduce the sum, hence we move $j$ -th pointer to the left . Otherwise, we increase $i$ -th pointer to the right to increase the sum.

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

1679 - Max Number of K-Sum Pairs (Medium)

Problem Link

Problem Statement

Approach 1: Hash Map

Approach 2: Two Pointers

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