add 3 sum solution

Author: BrianRuizy

15 - 3 Sum.md

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+# 15 - 3 Sum
+
+Difficulty: medium
+Done: No
+Last edited: February 22, 2022 1:23 AM
+Topic: array, sorting, two pointers
+
+## Problem
+
+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.
+
+## Solution
+
+Problem is framed like the popular two sum where we find 2 values $(x, y)$ which add up to given target, which can be solved by finding if difference between target and value $(target - x)$ exists in array. 
+
+However, here we are required to work with 3 values  $(x, y, z)$ and **find all triplets where sum of values is zero** $x+y+z=0$. This must be done without repeating the elements, though input array can contain duplicates.
+
+This approach will only work with sorted array
+
+### Pseudo
+
+1. Sort array
+2. Iterate through elements, setting current element as the pivot $x$. 
+    1. if current value (lowest) is greater than zero, then this would indicate there are only non-negative elements, which cannot sum to 0.
+    2. if current value is same as preceding (duplicate), then skip.
+    3. Proceed to step 3
+3. Use two pointer approach to compare values, starting with *left_ptr* at position $x+1$, and *right_ptr* at the end of the array or  $len(nums)$-1. We want to find all pairs whose sum is equal to 0, or $(x+y+z)=0$. 
+    
+    We will do so by creating separate function which essentially performs two-sum. Loop while *left_ptr* is smaller than *right_ptr*
+    
+    1. If $sum(x, y, z)$ is greater than 0, we will increment *left_ptr* 
+    2. Elif $sum(x,y, z)$ is less than 0, we will decrement *right_pt*
+    3. if sum is zero, meaning **triplet found,** add to results list. Increment left and decrement right. Increment left while while $left = left+1$, to skip duplicates.
+
+## Whiteboard
+
+[http://excalidraw.com](http://excalidraw.com)
+
+## Code
+
+```python
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        triplets = []
+        nums.sort()
+        
+        for i in range(len(nums)):
+            # check if all elements are non-negative
+            if nums[i] > 0:
+                break
+            
+            # skip duplicates
+            # call two_sum function to find triplets with i as target
+            if nums[i-1] != nums[i] or i == 0:
+                self.two_sum(i, nums, triplets)
+            
+        return triplets
+            
+        
+    def two_sum(self, i: int, nums: List[int], triplets: List[List[int]]):
+        # define left & right pointers
+        # loop while left & right pointers aren't at same position l < r
+        
+        left = i + 1
+        right = len(nums) - 1
+        
+        while left < right:
+            total = nums[i] + nums[left] + nums[right]
+
+            
+            #if sum > 0, increment left
+            if total < 0:
+                left += 1
+            
+            # if sum < 0, decrement right
+            elif total > 0: 
+                right -= 1
+            
+            else:
+                # if sum is 0, add to triplets list
+                triplets.append([nums[i], nums[left], nums[right]])
+                left += 1
+                right -= 1
+                
+                while left < right and nums[left] == nums[left-1]:
+                    # skip duplicates
+                    left += 1
+```