The Max Number of K-Sum Pairs problem from LeetCode 75 is a common interview question that tests your understanding of sorting, two-pointer techniques, and pair sum calculations. The challenge is simple to describe: given an array of numbers and a target sum k
, determine the maximum number of pairs whose sum equals k
. However, solving it efficiently for large arrays requires a smart approach.

Table of Contents
In this blog post, we’ll break down the problem, present a clean Python solution, and explain why it’s optimal for interviews and real-world applications.
Problem Statement
Given an integer array nums
and an integer k
, return the maximum number of operations you can perform on the array to form pairs whose sum equals k
. Each element can be used at most once.
Understanding the Problem
The key idea is to form pairs (nums[i], nums[j])
such that nums[i] + nums[j] = k
. Each number can only belong to one pair, so we need to be careful about counting.
Examples
Example 1:
Input: nums = [1,2,3,4], k = 5
Output: 2
Explanation: Pairs are (1,4) and (2,3)
Example 2:
Input: nums = [3,1,3,4,3], k = 6
Output: 1
Explanation: Only one pair sums to 6, e.g., (3,3)
Python Solution – Step-by-Step Explanation
Here’s a clean Python solution using the two-pointer technique:
class Solution:
def maxOperations(self, nums: List[int], k: int) -> int:
nums.sort()
if not nums:
return 0
start, end, result = 0, len(nums) - 1, 0
while start < end:
sum_pair = nums[start] + nums[end]
if sum_pair < k:
start += 1
elif sum_pair > k:
end -= 1
else:
start += 1
end -= 1
result += 1
return result
Step 1: Sort the Array
nums.sort()
- Sorting helps efficiently check sums from both ends.
- The smallest number is at
start
and the largest atend
.
Step 2: Initialize Two Pointers
start, end = 0, len(nums) - 1
result = 0
start
points to the smallest element.end
points to the largest element.result
keeps track of valid pairs found.
Step 3: Traverse and Count Pairs
while start < end:
sum_pair = nums[start] + nums[end]
if sum_pair < k:
start += 1
elif sum_pair > k:
end -= 1
else:
start += 1
end -= 1
result += 1
- If the sum is less than
k
, movestart
forward to increase the sum. - If the sum is greater than
k
, moveend
backward to decrease the sum. - If the sum equals
k
, a pair is found: incrementresult
and move both pointers.
Step 4: Return Maximum Pairs
return result
- After the pointers cross,
result
contains the maximum number of valid pairs.
Why This Solution Works
- Uses a two-pointer strategy to efficiently scan the sorted array.
- Ensures that each element is used at most once.
- Avoids nested loops, giving a linear time complexity after sorting.
Time and Space Complexity
Metric | Value |
---|---|
Time Complexity | O(n log n) – due to sorting |
Space Complexity | O(1) – only pointers and counters |
Edge Cases to Consider
Case | Output | Explanation |
---|---|---|
nums = [] | 0 | No pairs exist |
nums = [1,1], k = 2 | 1 | One valid pair |
nums = [1,2,3], k = 7 | 0 | No valid pair |
nums = [3,3,3,3], k = 6 | 2 | Handles duplicates correctly |
Real-World Applications
- Transaction matching: Pairing purchases and refunds efficiently.
- Resource allocation: Grouping items to meet target constraints.
- Interview prep: Tests understanding of greedy strategies and two-pointer techniques.
Conclusion
The Max Number of K-Sum Pairs problem is a classic example of how sorting combined with two-pointer techniques leads to efficient solutions. By carefully moving pointers and counting pairs, you can achieve optimal performance even on large arrays.
Mastering this problem not only strengthens your problem-solving skills for array-based challenges but also provides a strong foundation for tackling more complex pairing and summation problems in coding interviews and real-world tasks.
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- NLP Text Preprocessing Cheatsheet 2025: The Ultimate Powerful Guide
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External Links
🔗 LeetCode Problem Page – Max Number of K-Sum Pairs
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