class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        result = []
        
        nums.sort() 
        
        for i, a in enumerate(nums): 
            if i > 0 and a == nums[i - 1]: #this is so you don't reuse same value twice to avoid duplicates at end 
                continue 
            
            #next part is basically two sum 
            
            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: 
                    result.append([a, nums[l], nums[r]])
                    #only need to shift left pointer as right pointer will be moved by code above 
                    l += 1 
                    while nums[l] == nums[l - 1] and l < r: 
                        l += 1 
            
        return result 
      
      
#two sum solutions below for reference 
      
def twoSum(numbers, target): 
  left = 0 
  right = len(numbers) - 1 
  compare = 0 

  for i in range(0, len(numbers)): 
    compare = numbers[left] + numbers[right] 
    if compare > target: 
      right -= 1 
      elif compare < target: 
        left += 1 
        elif compare == target: 
          return [left + 1, right + 1] 

        
def twoSum(self, nums: List[int], target: int) -> List[int]:
  num_dictionary = {}
  for i in range(0, len(nums)):
    number = nums[i]
    if target - number not in num_dictionary: 
      #number will be key and index (i) will be value 
      #adding number to dictionary AFTER checking 1st value 
      #if [2, 1, 3] and target = 4, 4 -2 = 2 and 2 will be present in dictionary but we cannot use 2 + 2 
      num_dictionary[number] = i 
      elif target - number in num_dictionary: 
        return [i, num_dictionary[target - number]]