What is the brute force approach and complexity for Max Number of K-Sum Pairs?

For each unused number, scan for an unused partner; mark both used with time complexity O(N²) and space complexity O(N).

What is the optimal approach and complexity for Max Number of K-Sum Pairs?

If cnt[k − x] > 0, pair and decrement; else cnt[x]++ with optimal time complexity O(N) and space complexity O(N).

Max Number of K-Sum Pairs

Q: How do you identify the pattern for Max Number of K-Sum Pairs?

Look for key signals in the prompt: "two numbers whose sum equals k" indicates The partner of `x` is fixed: `k − x`. Complement lookup again.; "remove them from the array" indicates Each number is used once. The map must track how many copies are **left**, not just whether a value was seen.; "maximum number of operations" indicates Count disjoint pairs. Pairing greedily works because copies of the same value are interchangeable..

Medium·LeetCode #1679 Pro OnlySlight Variation

LeetCode

Problem Statement

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

In one operation you pick two numbers from the array whose sum is exactly k and remove both of them. A removed number can't be used again.

Return the largest number of operations you can do.

Input

An integer array nums and an integer k.

Output

The maximum number of operations, i.e. the most disjoint pairs that each sum to k.

Example 1

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

Output: 2

Remove 1 and 4, then remove 2 and 3. Nothing is left.

Example 2

Input: nums = [3, 1, 3, 4, 3], k = 6

Output: 1

Remove two of the 3s. The third 3 has no partner left, and 1 and 4 have no partners at all.

Constraints

• 1 ≤ nums.length ≤ 10^5
• 1 ≤ nums[i] ≤ 10^9
• 1 ≤ k ≤ 10^9

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“two numbers whose sum equals k”

→ The partner of `x` is fixed: `k − x`. Complement lookup again.

“remove them from the array”

→ Each number is used once. The map must track how many copies are **left**, not just whether a value was seen.

“maximum number of operations”

→ Count disjoint pairs. Pairing greedily works because copies of the same value are interchangeable.

Approach Complexity Ladder

Brute Force — find a partner for each numberO(N²)O(N)

For each unused number, scan for an unused partner; mark both used

Better — sort + two pointersO(N log N)O(1)

Sort, then pair the ends: equal → take both, too small → left++, too big → right--

Optimal — count map of waiting numbersO(N)O(N)

If cnt[k − x] > 0, pair and decrement; else cnt[x]++

Brute Force Intuition

Keep a used flag for every position. For each unused nums[i], scan the rest of the array for an unused nums[j] with nums[i] + nums[j] == k. If found, mark both used and count one operation.

Recognition rule for this sub-pattern:

For each element, ask "what partner do I need?" and check if that partner is already in the map.

Remember this · Arrays & Hashing

Trigger

"Have I seen X before?", pairs that add up to a target, grouping things that look alike, counting occurrences.

Think

Instead of scanning again for each element, can I remember earlier elements in a map?

Key question

What exactly do I store as the key, and what as the value?

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Contains Duplicate II