Problem: 3Sum - Finding Unique Triplets

Problem Statement

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.

The solution set must not contain duplicate triplets.

Approach: Sorting + Two Pointers

We can reduce the 3Sum problem to a "Two Sum" problem by fixing one number and searching for the other two.

1. Sort the array: This allows us to use two pointers and easily skip duplicates.
2. Iterate: Fix nums[i] and look for a pair in nums[i+1...n-1] that sums to -nums[i].
3. Skip Duplicates: To avoid duplicate triplets, skip the current i if nums[i] == nums[i-1]. Also, skip duplicates when moving the two pointers.

Java Implementation

public List<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();
    
    for (int i = 0; i < nums.length - 2; i++) {
        // Skip duplicate fixed element
        if (i > 0 && nums[i] == nums[i - 1]) continue;
        
        int left = i + 1;
        int right = nums.length - 1;
        int target = -nums[i];
        
        while (left < right) {
            int sum = nums[left] + nums[right];
            if (sum == target) {
                res.add(Arrays.asList(nums[i], nums[left], nums[right]));
                // Skip duplicates for left and right
                while (left < right && nums[left] == nums[left + 1]) left++;
                while (left < right && nums[right] == nums[right - 1]) right--;
                left++;
                right--;
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
        }
    }
    return res;
}

Complexity Discussion

• Time Complexity: $O(n^2)$. Sorting takes $O(n \log n)$, and the nested loops take $O(n^2)$.
• Space Complexity: $O(\log n)$ to $O(n)$ depending on the sorting implementation.

Interview Tips

• Why sort? "Sorting helps in two ways: it enables the two-pointer approach and makes it easy to identify and skip duplicate elements."
• Edge cases: Array length less than 3, all zeros, no valid triplets.