15. 3Sum
mediumAsked at AMDFind all unique triplets in an array that sum to zero. AMD uses this to test whether candidates can extend a two-pointer pattern and handle deduplication without a set — skills that transfer to collision detection and register-bank conflict resolution in compiler backends.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Source citations
Public interview reports confirming this problem appears in AMD loops.
- Glassdoor (2026-Q1)— AMD SWE onsite candidates report 3Sum as a core medium problem appearing in the second coding round.
- Blind (2025-10)— AMD interview prep threads list 3Sum as a two-pointer/sorting medium commonly asked in SWE roles.
Problem
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.
Constraints
3 <= nums.length <= 3000−10^5 <= nums[i] <= 10^5
Examples
Example 1
nums = [-1,0,1,2,-1,-4][[-1,-1,2],[-1,0,1]]Explanation: Two unique triplets sum to zero.
Example 2
nums = [0,0,0][[0,0,0]]Explanation: Only one unique triplet.
Approaches
1. Sort + Two Pointers
Sort the array. Fix the first element with an outer loop, then use two pointers (left=i+1, right=end) to find pairs summing to -nums[i]. Skip duplicates at each level.
- Time
- O(n^2)
- Space
- O(1) excluding output
function threeSum(nums) {
nums.sort((a, b) => a - b);
const result = [];
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue; // skip duplicate i
if (nums[i] > 0) break; // sorted: no pair can cancel a positive pivot
let left = i + 1, right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === 0) {
result.push([nums[i], nums[left], nums[right]]);
while (left < right && nums[left] === nums[left + 1]) left++;
while (left < right && nums[right] === nums[right - 1]) right--;
left++; right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return result;
}Tradeoff: O(n^2) time — optimal for this problem (proven lower bound). O(1) extra space. The deduplication skip loops are the tricky part; walk through them carefully.
AMD-specific tips
AMD interviewers appreciate candidates who articulate why O(n^2) is optimal here: the output alone can be O(n^2) triplets, so you can't do better. Explain the three deduplication points: skip repeated pivot (outer loop), skip repeated left after finding a match, skip repeated right after finding a match. Early exit when nums[i] > 0 is a micro-optimization that signals performance instinct.
Common mistakes
- Forgetting to skip duplicate pivots in the outer loop — produces duplicate triplets.
- Skipping duplicates before advancing left and right — you skip past the match itself.
- Using a Set of arrays for deduplication — JS arrays don't compare by value in Sets.
- Forgetting the early break when nums[i] > 0 — correct but less efficient.
Follow-up questions
An interviewer at AMD may pivot to one of these next:
- 4Sum (LC 18) — generalize the outer loop to two fixed indices, still O(n^3).
- 3Sum Closest (LC 16) — find the triplet whose sum is closest to target.
- How would you parallelize 3Sum across GPU threads for a very large array?
Solve it now
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FAQ
Why sort first?
Sorting enables the two-pointer approach (shrink the window based on whether the sum is too small or too large) and makes deduplication trivial via adjacent-element comparison.
Why can we break when nums[i] > 0?
After sorting, nums[left] >= nums[i] > 0 and nums[right] >= nums[i] > 0, so the sum is always positive. No valid triplet is possible.
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