20. 3Sum
mediumAsked at UdemyFind all unique triplets that sum to zero — Udemy uses this to test deduplication logic relevant to course bundle pricing and recommendation filtering.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i, j, k are distinct and nums[i] + nums[j] + nums[k] == 0. 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]]Example 2
nums = [0,0,0][[0,0,0]]Approaches
1. Brute force
Try every triple with three nested loops; use a Set to deduplicate — O(n^3).
- Time
- O(n^3)
- Space
- O(n)
// three nested loops + deduplication via sorted key in Set
// correct but too slow for n=3000Tradeoff:
2. Sort + two pointers
Sort the array; for each index i fix nums[i] as the first element and use a left/right pointer pair to find complementary pairs; skip duplicates at both the outer and inner loop levels.
- Time
- O(n^2)
- Space
- O(1)
function threeSum(nums) {
nums.sort((a, b) => a - b);
const res = [];
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i-1]) continue;
let l = i+1, r = nums.length-1;
while (l < r) {
const sum = nums[i] + nums[l] + nums[r];
if (sum === 0) {
res.push([nums[i], nums[l], nums[r]]);
while (l < r && nums[l] === nums[l+1]) l++;
while (l < r && nums[r] === nums[r-1]) r--;
l++; r--;
} else if (sum < 0) l++;
else r--;
}
}
return res;
}Tradeoff:
Udemy-specific tips
Udemy asks about e-learning recommendation systems, content search, and marketplace algorithms — balanced mix of arrays, hash maps, and dynamic programming problems.
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