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13. 3Sum

mediumAsked at Adyen

Find all unique triplets in the array that sum to zero.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given an integer array nums, return all unique triplets [a, b, c] such that a + b + c == 0. The solution set must not contain duplicate triplets.

Constraints

  • 3 <= nums.length <= 3000
  • -10^5 <= nums[i] <= 10^5

Examples

Example 1

Input
nums = [-1,0,1,2,-1,-4]
Output
[[-1,-1,2],[-1,0,1]]

Example 2

Input
nums = [0,1,1]
Output
[]

Approaches

1. Brute force triple loop

Try every triple, dedupe with a set.

Time
O(n^3)
Space
O(n)
const out = new Set();
for (let i = 0; i < nums.length; i++)
  for (let j = i+1; j < nums.length; j++)
    for (let k = j+1; k < nums.length; k++)
      if (nums[i] + nums[j] + nums[k] === 0) out.add([nums[i], nums[j], nums[k]].sort().join(','));
return [...out].map(s => s.split(',').map(Number));

Tradeoff:

2. Sort + two-pointer

Sort, fix an anchor, two-pointer-sweep the remainder.

Time
O(n^2)
Space
O(1)
function threeSum(nums) {
  nums.sort((a,b) => a-b);
  const out = [];
  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 s = nums[i] + nums[l] + nums[r];
      if (s === 0) { out.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 (s < 0) l++;
      else r--;
    }
  }
  return out;
}

Tradeoff:

Adyen-specific tips

Adyen interviewers look for the duplicate-skipping discipline — payment reconciliation rejects double-counted triplets the same way and they grade the dedupe carefully.

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