3Sum.py

# 3Sum

# 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.


# Answer
def threeSum(self, nums):
    nums.sort()  
    result = []
    n = len(nums)
    
    for i in range(n - 2):  
        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]])
                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