Solved, with help from solutions

NeetCodeRoadmap/Arrays_and_Hashing/15_3Sum.py

@@ -4,14 +4,52 @@ def threeSum(nums):
     :rtype: List[List[int]]
     """
     # Always end at net zero from the sum
-    first = 0
-    second = 1
-    solutionSets = []
+    # first = 0
+    # endpoint = len(nums)-1
+    # solutionSets = []
 
-    for i in range(len(nums)):
-        if nums[first] + nums[second] + nums[i] == 0:
-            solutionSets.append([nums[first], nums[second], nums[i]])
+    # # Handle edge-case of small list with only 3 variables
+    # if len(nums) == 3 and sum(nums) == 0:
+    #     solutionSets.append(nums)
+    #     return solutionSets
 
+    # while first != len(nums)-2:
+    #     for i in range(first+1, len(nums)):
+    #         if nums[first] + nums[i] + nums[endpoint] == 0 and [nums[first], nums[i], nums[endpoint]] not in solutionSets:
+    #             solutionSets.append([nums[first], nums[i], nums[endpoint]])
+    #         else:
+    #             endpoint -= 1
+    #     first += 1
+    # return solutionSets
 
+    # Solution
+    nums.sort()
+    answer = [] 
+
+    # Loop through until the last triplet, solution relies on negative indexes
+    for i in range(len(nums) - 2): 
+        if nums[i] > 0: # end the loop if the values are positive
+            break
+        if i > 0 and nums[i] == nums[i-1]: # skip the current i index if it's the same as the last one
+            continue
+        l = i + 1 # Begin the left index at i + 1
+        r = len(nums) - 1 # Begin the right index at the end
+        while l < r:
+            total = nums[i] + nums[l] + nums[r]
+            if total < 0: # If the total is negative, we increase the negative side to get closer to 0
+                l += 1
+            elif total > 0: # If the total is positive, we decrease the positive side to get closer to 0
+                r -= 1
+            else:
+                triplet = [nums[i], nums[l], nums[r]]
+                answer.append(triplet)
+                while l < r and nums[l] == triplet[1]: # Continue increasing the left index until a different 
+                    l += 1 # number is found to avoid duplicates
+                while l < r and nums[r] == triplet[2]: # Same with decreasing the right index
+                    r -= 1
+    return answer
     
 print(threeSum([-1,0,1,2,-1,-4]))
+print(threeSum([0,0,0]))
+print(threeSum([0,0,0,0]))
+print(threeSum([1,2,-2,-1]))