Published Nov 29, 2021
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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.
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Input: nums = []
Output: []
Input: nums = [0]
Output: []
- 0 <= nums.length <= 3000
- -105 <= nums[i] <= 105
function threeSum(nums: number[]): number[][] {
const res = []
// notice how we sort the array here
nums.sort((a, b) => a - b)
// we use i as the first element in result
for(let i = 0; i < nums.length - 2; i++) {
// we dont need any duplicate first elements
if(i > 0 && nums[i] === nums[i - 1]) continue;
// set index for second & last element
let l = i + 1, r = nums.length - 1;
// move to mid
while(l < r) {
const sum = nums[i] + nums[l] + nums[r]
if(sum > 0){
r--;
} else if (sum < 0) {
l++;
} else {
res.push([nums[i], nums[l], nums[r]])
l++
while (nums[l] === nums[l - 1] && l < r) l++;
}
}
}
return res;
};