solve 3sum

Author: SamTheKorean

3sum/samthekorean.py

@@ -0,0 +1,29 @@
+# O(n^2) where there is nested loop.
+# O(1) where high and low are reused as constant values.
+
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        # Initialize a set to store unique triplets
+        triplets = set()
+
+        # Sort the list to make two-pointer approach possible
+        nums.sort()
+
+        # Iterate through the list, considering each element as a potential first element of a triplet
+        for i in range(len(nums) - 2):
+            low = i + 1
+            high = len(nums) - 1
+
+            # To avoid duplicate iteration
+            while low < high:
+                three_sum = nums[i] + nums[low] + nums[high]
+
+                if three_sum < 0:
+                    low += 1
+                elif three_sum > 0:
+                    high -= 1
+                else:
+                    triplets.add((nums[i], nums[low], nums[high]))
+                    low += 1
+                    high -= 1