Leetcode•Oct 15, 2025
3Sum
Hazrat Ali
Leetcode
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.
Example 1:
Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.
Solution
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(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 left = i + 1;
let right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === 0) {
res.push([nums[i], nums[left], nums[right]]);
left++;
right--;
while (left < right && nums[left] === nums[left - 1]) left++;
while (left < right && nums[right] === nums[right + 1]) right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return res;
};
console.log(threeSum([-1, 0, 1, 2, -1, -4]));
console.log(threeSum([0, 1, 1]));
console.log(threeSum([0, 0, 0]));