3Sum

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Problem

• We’re given an input array of numbers, our function should return all the triplets [nums[i], nums[j], nums[k]] where [nums[i], nums[j], nums[k]] == 0.
• The indices i, j, and k are all distinct.

Example

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

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Approach 1

Visualize the problem

Solve without code

• Sort the values
• Starting from the left to right we can find combinations by using the starting number and combining the left and right of the input array
• If the sum is too large we can reduce the right pointer, conversely we increase the left pointer
• Once we’re done with the possible combinations for this current pointer a we can move on to the next

Solve with code

• Sort the array
• Create two loops
• Define a current index a and left and right pointers band c. a will act as our outer loop and b and c will essentially be two sum within our outer loop.
• Increase or decrease b and c based on the sum
• Once we’re done with the current index a we can increase the index by one and reset b and cand repeat the process.

Implementation

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        # Sort input array
        nums.sort()
        res = []
        
        for i, a in enumerate(nums):
            if i > 0 and a == nums[i - 1]:
                continue
            l, r = i + 1, len(nums) - 1
            while l < r:
                threeSum = a + nums[l] + nums[r]
                if threeSum > 0:
                    r -= 1
                elif threeSum < 0:
                    l += 1
                else:
                    res.append([a, nums[l], nums[r]])
                    l += 1
                    while nums[l] == nums[l-1] and l < r:
                        l+= 1
        
        return res

Complexity Analysis

• Time Complexity: $O(nlogn)+O(n^2)=O(n^2)$ - Sorting and Two loops.
• Space Complexity: $O(1)$