How do you solve Max Number of K-Sum Pairs on LeetCode?

Max Number of K-Sum Pairs (LeetCode #1679) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map can significantly reduce the time complexity.

What is the brute force solution for Max Number of K-Sum Pairs?

The brute force approach for Max Number of K-Sum Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible pair of numbers in the array to see if their sum equals k. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Max Number of K-Sum Pairs?

Max Number of K-Sum Pairs uses the following concepts: Array, Hash Table, Two Pointers, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Max Number of K-Sum Pairs?

Always think about how to optimize a brute-force solution. Explain your thought process clearly while coding. Consider edge cases, such as duplicates and pairs that are the same.

What are common mistakes when solving Max Number of K-Sum Pairs?

Not considering the case where the number and its complement are the same. Failing to update the frequency count correctly after forming pairs.

#1679

Max Number of K-Sum Pairs

Medium

ArrayHash TableTwo PointersSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a hash map to count occurrences of each number. This allows us to efficiently find pairs that sum to k without needing nested loops.

⚙️

Algorithm

4 steps

1Step 1: Create a hash map to count occurrences of each number in the array.
2Step 2: Iterate through each number in the array and calculate its complement (k - number).
3Step 3: Check if the complement exists in the hash map. If it does, calculate the number of pairs that can be formed and update the count.
4Step 4: Decrease the count of the current number and its complement in the hash map to avoid reusing them.

solution.py17 lines

1# Full working Python code
2
3def maxOperations(nums, k):
4    count = 0
5    freq = {}
6    for num in nums:
7        freq[num] = freq.get(num, 0) + 1
8    for num in nums:
9        complement = k - num
10        if complement in freq:
11            pairs = min(freq[num], freq[complement])
12            if num == complement:
13                pairs //= 2
14            count += pairs
15            freq[num] -= pairs
16            freq[complement] -= pairs
17    return count

ℹ

Complexity note: The time complexity is O(n) because we make a single pass to count frequencies and another pass to find pairs. The space complexity is O(n) due to the hash map storing counts of each number.

1Using a hash map can significantly reduce the time complexity.
2Pairs can be formed by checking complements of each number.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.