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threeSum.js
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58 lines (47 loc) · 1.5 KB
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/**
* LeetCode 15. 3Sum
* https://leetcode.com/problems/3sum/
*
* Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]]
* such that i != j, i != k, j != k, and nums[i] + nums[j] + nums[k] == 0.
*
* The solution set must not contain duplicate triplets.
*/
/**
* @param {number[]} nums
* @return {number[][]}
*/
function threeSum(nums) {
const results = [];
// 1. Sort the array numerically
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length - 2; i++) {
// Optimization: If the current number is > 0, the sum of
// three positive numbers can never be 0.
if (nums[i] > 0) break;
// 2. Skip duplicates for the first element
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) {
results.push([nums[i], nums[left], nums[right]]);
// 3. Skip duplicates for the second and third elements
while (left < right && nums[left] === nums[left + 1]) left++;
while (left < right && nums[right] === nums[right - 1]) right--;
// Move both pointers after finding a match
left++;
right--;
} else if (sum < 0) {
// Sum is too low, move left pointer to a larger number
left++;
} else {
// Sum is too high, move right pointer to a smaller number
right--;
}
}
}
return results;
}
module.exports = { threeSum };