# Solving Leetcode Interviews in Seconds with AI: Max Number of K-Sum Pairs


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1679" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 <= 105 1 <= nums[i] <= 109 1 <= k <= 109  

	# Explanation
	Here's an efficient solution to find the maximum number of operations:

*   **Core Idea:** Use a hash map (or dictionary in Python) to store the frequency of each number in the input array. Iterate through the array, and for each number `num`, check if `k - num` exists in the hash map.
*   **Counting Operations:** If `k - num` exists, decrement the counts of both `num` and `k - num` in the hash map. If the counts are greater than 0 after decrementing (meaning there are still instances of these numbers available), increment the operation count.
*   **Optimization:** This approach avoids unnecessary iterations and efficiently finds pairs that sum to `k`.

*   **Complexity:** Runtime: O(n), where n is the length of the input array. Storage: O(n) in the worst case, to store the frequency of each number in the hash map.

	
	# Code
	```python
	def maxOperations(nums: list[int], k: int) -> int:
    """
    Given an integer array nums and an integer k, return the maximum number of operations you can perform on the array.
    In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.
    """
    counts = {}
    operations = 0
    for num in nums:
        counts[num] = counts.get(num, 0) + 1

    for num in nums:
        complement = k - num
        if counts[num] > 0 and complement in counts and counts[complement] > 0:
            if num == complement and counts[num] < 2:
                continue
            counts[num] -= 1
            counts[complement] -= 1
            operations += 1
            
    return operations
	```