LeetCode #1679 — MEDIUM

Max Number of K-Sum Pairs

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Two Pointers

Example 1

[1,2,3,4]
5

Example 2

[3,1,3,4,3]
6

Core Insight

What unlocks the optimal approach

The abstract problem asks to count the number of disjoint pairs with a given sum k.
For each possible value x, it can be paired up with k - x.
The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1679: Max Number of K-Sum Pairs
class Solution {
    public int maxOperations(int[] nums, int k) {
        Arrays.sort(nums);
        int l = 0, r = nums.length - 1;
        int ans = 0;
        while (l < r) {
            int s = nums[l] + nums[r];
            if (s == k) {
                ++ans;
                ++l;
                --r;
            } else if (s > k) {
                --r;
            } else {
                ++l;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1679: Max Number of K-Sum Pairs
func maxOperations(nums []int, k int) int {
	sort.Ints(nums)
	l, r, ans := 0, len(nums)-1, 0
	for l < r {
		s := nums[l] + nums[r]
		if s == k {
			ans++
			l++
			r--
		} else if s > k {
			r--
		} else {
			l++
		}
	}
	return ans
}

# Accepted solution for LeetCode #1679: Max Number of K-Sum Pairs
class Solution:
    def maxOperations(self, nums: List[int], k: int) -> int:
        nums.sort()
        l, r, ans = 0, len(nums) - 1, 0
        while l < r:
            s = nums[l] + nums[r]
            if s == k:
                ans += 1
                l, r = l + 1, r - 1
            elif s > k:
                r -= 1
            else:
                l += 1
        return ans

// Accepted solution for LeetCode #1679: Max Number of K-Sum Pairs
impl Solution {
    pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
        let mut nums = nums.clone();
        nums.sort();
        let (mut l, mut r, mut ans) = (0, nums.len() - 1, 0);
        while l < r {
            match nums[l] + nums[r] {
                sum if sum == k => {
                    ans += 1;
                    l += 1;
                    r -= 1;
                }
                sum if sum > k => {
                    r -= 1;
                }
                _ => {
                    l += 1;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1679: Max Number of K-Sum Pairs
function maxOperations(nums: number[], k: number): number {
    const cnt = new Map();
    let ans = 0;
    for (const x of nums) {
        if (cnt.get(k - x)) {
            cnt.set(k - x, cnt.get(k - x) - 1);
            ++ans;
        } else {
            cnt.set(x, (cnt.get(x) | 0) + 1);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.