15. 3Sum Leetcode Solution

The “3Sum” problem is a classic algorithmic challenge where the goal is to find all unique triplets in an array that sum up to a target value. In this blog post, we will delve into three Python solutions for the 3Sum problem. Each solution will be thoroughly explained, including its steps and a comprehensive analysis of time and space complexity.

Solution 1: Brute Force

def three_sum_bruteforce(nums):
    result = []

    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            for k in range(j + 1, len(nums)):
                if nums[i] + nums[j] + nums[k] == 0:
                    triplet = [nums[i], nums[j], nums[k]]
                    triplet.sort()
                    if triplet not in result:
                        result.append(triplet)

    return result

Explanation:

1. Triple Nested Loop: The brute force solution utilizes three nested loops to iterate through all possible triplets in the array.
2. Check and Append: For each triplet, it checks whether the sum is equal to zero, sorts the triplet, and appends it to the result list if it is not already present.

Time Complexity: O(n³)

• The time complexity is cubic due to the triple nested loops, making it highly inefficient for large input arrays.

Space Complexity: O(1)

• The space complexity is constant as the solution uses only a few variables regardless of the input size.

Solution 2: Two-Pointer Approach

def three_sum_two_pointer(nums):
    nums.sort()
    result = []

    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue

        target = -nums[i]
        left, right = i + 1, len(nums) - 1

        while left < right:
            current_sum = nums[left] + nums[right]

            if current_sum == target:
                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 current_sum < target:
                left += 1
            else:
                right -= 1

    return result

Explanation:

1. Sorting: The array is sorted to facilitate the two-pointer approach.
2. Two-Pointer Technique: The solution employs two pointers, left and right, moving towards each other to find triplets that sum to the target.

Time Complexity: O(n²)

• The time complexity is quadratic, as the two-pointer approach involves a single loop through the array.

Space Complexity: O(1)

• The space complexity is constant as the solution uses only a few variables regardless of the input size.

Solution 3: Hash Set

def three_sum_hash_set(nums):
    nums.sort()
    result = set()

    for i in range(len(nums) - 2):
        target = -nums[i]
        seen = set()

        for j in range(i + 1, len(nums)):
            complement = target - nums[j]

            if complement in seen:
                result.add((nums[i], complement, nums[j]))

            seen.add(nums[j])

    return list(result)

Explanation:

1. Sorting: Similar to the two-pointer approach, the array is sorted to simplify the search.
2. Hash Set: A set (seen) is used to store the elements encountered during the iteration.

Time Complexity: O(n²)

• The time complexity is quadratic, as the solution involves a nested loop through the array.

Space Complexity: O(n)

• The space complexity is linear due to the use of the seen set, which can potentially store all elements in the array.

Conclusion:

In this blog post, we explored three Python solutions for the “3Sum” problem, each employing a distinct strategy. The brute force approach, while simple, suffers from cubic time complexity, making it impractical for larger datasets. The two-pointer approach and hash set solution offer more efficient alternatives with quadratic time complexity. Understanding these solutions provides valuable insights into algorithmic problem-solving, enabling the selection of the most appropriate approach based on the specific problem constraints.