3Sum

Problem Statement

Given an array of integers, find all unique triplets that sum to zero. This is a medium-level problem combining sorting with two pointers.

Example

Input: nums = [-1, 0, 1, 2, -1, -4]

Output: [[-1, -1, 2], [-1, 0, 1]] (unique triplets summing to zero)

Code

public class Solution {
    public List> threeSum(int[] nums) {
        Arrays.sort(nums);
        List> result = new ArrayList<>();
        
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            
            int left = i + 1, right = nums.length - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0) {
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    while (left < right && nums[right] == nums[right - 1]) right--;
                    left++;
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }
}
            

def three_sum(nums):
    nums.sort()
    result = []
    
    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue
        
        left, right = i + 1, len(nums) - 1
        while left < right:
            curr_sum = nums[i] + nums[left] + nums[right]
            if curr_sum == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1
                left += 1
                right -= 1
            elif curr_sum < 0:
                left += 1
            else:
                right -= 1
    return result

# Example usage
print(three_sum([-1, 0, 1, 2, -1, -4]))  # [[-1, -1, 2], [-1, 0, 1]]
            

function threeSum(nums) {
    nums.sort((a, b) => a - b);
    const result = [];
    
    for (let i = 0; i < nums.length - 2; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;
        
        let left = i + 1, right = nums.length - 1;
        while (left < right) {
            let sum = nums[i] + nums[left] + nums[right];
            if (sum === 0) {
                result.push([nums[i], nums[left], nums[right]]);
                while (left < right && nums[left] === nums[left + 1]) left++;
                while (left < right && nums[right] === nums[right - 1]) right--;
                left++;
                right--;
            } else if (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

// Example usage
console.log(threeSum([-1, 0, 1, 2, -1, -4])); // [[-1, -1, 2], [-1, 0, 1]]
            

Explanation

• Sort the array to handle duplicates and enable two-pointer approach.
• Fix one element and use two pointers to find the other two.
• If sum is zero, add triplet to result and skip duplicates.
• Adjust pointers based on whether sum is less than or greater than zero.

Note