# Solving Leetcode Interviews in Seconds with AI: 3Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "15" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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. 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.    Constraints:  3 <= nums.length <= 3000 -105 <= nums[i] <= 105  

	# Explanation
	Here's the breakdown of the solution and the Python code:

*   **High-Level Approach:**
    *   Sort the input array to efficiently use the two-pointer technique.
    *   Iterate through the array, fixing one number and then using two pointers to find the other two numbers that sum to the negative of the fixed number.
    *   Skip duplicate numbers to avoid duplicate triplets in the result.

*   **Complexity:**
    *   Runtime: O(n^2), where n is the length of the input array. Sorting takes O(n log n), but the nested loops dominate.
    *   Storage: O(1) or O(n) depending on sorting algorithm (in-place or not). The space used for the result list is not considered extra space.

	
	# Code
	```python
	def threeSum(nums):
    nums.sort()
    result = []
    n = len(nums)

    for i in range(n - 2):
        # Skip duplicate elements for the first number
        if i > 0 and nums[i] == nums[i - 1]:
            continue

        left = i + 1
        right = n - 1

        while left < right:
            total = nums[i] + nums[left] + nums[right]

            if total == 0:
                result.append([nums[i], nums[left], nums[right]])

                # Skip duplicate elements for the second and third numbers
                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 total < 0:
                left += 1
            else:
                right -= 1

    return result
	```