15. 3Sum
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.
/**
* @param {number[][]} matrix
* @return {number[][]}
*/
function threeSum(input: number[]) {
const length = input.length;
input.sort((a, b) => a - b);
const res = new Set<string>();
function checkPosition(mid: number) {
let left = mid - 1;
let right = mid + 1;
while (left >= 0 && right < length) {
const sum = input[left] + input[mid] + input[right];
if (sum === 0) {
res.add([input[left], input[mid], input[right]].join(","));
left--;
}
if (sum < 0) {
right++;
}
if (sum > 0) {
left--;
}
}
}
for (let i = 1; i < length - 1; i++) {
checkPosition(i);
}
return [...res].map((v) => v.split(","));
}