#1679 - Max Number of K-Sum Pairs

MediumTwo Pointers

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. Return the maximum number of such operations you can perform.

Input	Output
nums = [1,2,3,4], k = 5	2
nums = [3,1,3,4,3], k = 6	1
nums = [2,2,2,2], k = 4	2

Solution Space

Try it

/**
 * Counts the maximum number of pairs whose sum equals k.
 *
 * @param nums Array of positive integers
 * @param k Target pair sum
 * @returns Maximum number of k-sum pair removals
 */
export function maxOperations(nums: number[], k: number): number {
  const freq = new Map<number, number>();
  let ops = 0;

  for (const num of nums) {
    const complement = k - num;

    if ((freq.get(complement) ?? 0) > 0) {
      // Complement available — form a pair
      ops++;
      freq.set(complement, freq.get(complement)! - 1);
    } else {
      freq.set(num, (freq.get(num) ?? 0) + 1);
    }
  }

  return ops;
}

Approach: hash map (single pass) – count frequencies of elements and find pairs that sum to k, decrementing counts as pairs are found.

Time: O(n) – one pass through the array with O(1) map operations.

Space: O(n) – worst-case every element is stored in the frequency map.

Alternative approaches:

Approach	Idea	Time	Space
Sort + two pointers	Sort the array, then use left/right pointers moving inward: if sum < k move left++, if sum > k move right--, if equal count and move both.	O(n log n)	O(1) (in-place sort)

Why the hash-map method is preferred:

• Achieves O(n) time in a single pass without mutating the input.
• The sort + two-pointer approach is a solid alternative when O(1) extra space is more important than time.

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