Journal Archive

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
    const result = [];
    
    // sort number for easier comparison
    nums.sort((a, b) => a - b);
    
    for (let i = 0; i < nums.length - 2; i++) {
        // skip same leading number
        if (i === 0 || i > 0 && nums[i] !== nums[i - 1]) {
            let left = i + 1;
            let right = nums.length - 1;
            
            // two pointer approach
            while (left < right) {
                let sum = nums[i] + nums[left] + nums[right];
                if (sum < 0) {
                    left++;
                }
                else if (sum > 0) {
                    right--;
                }
                else {
                    result.push([nums[i], nums[left], nums[right]]);
                    // skip duplicates
                    
                    while (left < right && nums[left] === nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] === nums[right - 1]) {
                        right--;
                    }
                    
                    left++;
                    right--;
                }
            } 
        }
 
    }
    
    return result;
};

Day 77: Solving one of LeetCode problems

15. 3Sum Difficulty - Medium

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]]
		

Example 2:

Input: nums = []
Output: []
		

Example 3:

Input: nums = [0]
Output: []