Introduction to the 3Sum Problem

The 3Sum (LeetCode 15) problem is a classic algorithm challenge often encountered in software engineering interviews, particularly at companies like Google, Facebook, and Amazon.

The task is straightforward yet tricky: given an array of integers, find all unique triplets in the array that add up to zero.

Specifically, the requirement is that each triplet [nums[i], nums[j], nums[k]] must satisfy the conditions where i, j, and k are distinct indices, and nums[i] + nums[j] + nums[k] == 0.

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.

This problem not only tests your ability to manipulate arrays but also your understanding of optimization strategies in reducing time complexity.

Exploring Different Approaches

1. Brute Force: The most intuitive approach is the brute force method, where three nested loops would check every possible triplet combination. However, this results in a prohibitive time complexity of O(n^3), which is inefficient for larger arrays.

2. Hash Table: By using a hash table to store numbers, you can potentially reduce the complexity to O(n^2) by looking up the complement of two numbers currently being considered. However, handling duplicates and ensuring the indices meet the problem's requirements complicate this approach.

3. Sorting with Two-pointers: This is the most efficient approach. By sorting the array first, you can use a two-pointer technique to find pairs that sum to the negative of the current number. This approach efficiently avoids duplicates and runs in O(n^2), which is the best achievable time complexity for this problem.

Why Avoiding Duplicates is Crucial

Avoiding duplicates is critical because the problem specifies that the solution set must contain only unique triplets.

Including duplicates would not only fail to meet the problem's requirements but also significantly increase the output size, which could lead to performance issues on larger datasets.

Step-by-Step Solution

• Sort the Array: Begin by sorting the array. This helps in efficiently finding pairs and skipping duplicates.

• Iterate with a Fixed Pointer: Fix one number at a time and use a two-pointer approach for the rest of the array.

• Apply Two-pointer Technique: For each fixed number, initialize two pointers—one at the start (right after the fixed pointer) and one at the end of the array. Move these pointers based on the sum comparison.

• Check for Zero Sum: If the sum of the numbers at the two pointers with the fixed number is zero, record the triplet.

• Skip Duplicates: After finding a triplet or moving a pointer, always skip the duplicate numbers to avoid duplicate triplets in the result.

Time Complexity: The overall time complexity of this solution is O(n^2), due to the sorting step and the subsequent two-pointer sweep of the array.

Python Solution

def three_sum(nums):
    nums.sort()
    results = []
    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue  # Skip duplicate fixed numbers
        left, right = i + 1, len(nums) - 1
        while left < right:
            current_sum = nums[i] + nums[left] + nums[right]
            if current_sum == 0:
                results.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1  # Skip duplicate left numbers
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1  # Skip duplicate right numbers
                left += 1
                right -= 1
            elif current_sum < 0:
                left += 1
            else:
                right -= 1
    return results

Conclusion

The 3Sum problem is an excellent test of both your understanding of array manipulation and optimization strategies.

The optimal solution leverages sorting and a two-pointer technique, balancing efficiency with the ability to handle duplicates effectively.

Understanding and implementing these strategies can greatly enhance your performance in coding interviews and your overall problem-solving skills in competitive programming and real-world software development.

This comprehensive guide aims to fully equip you with the knowledge to tackle the 3Sum problem efficiently, ensuring you stand out in your next coding interview.