feat: solve 3sum

Author: lylaminju

3sum/pmjuu.py

@@ -0,0 +1,42 @@
+from typing import List
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        result = []
+        nums.sort() # sort nums before using two-pointers
+        
+        for i, num in enumerate(nums):
+            # skip duplicated targets
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+
+            target = -num
+            left, right = i + 1, len(nums) - 1
+            
+            while left < right:
+                if nums[left] + nums[right] == target:
+                    result.append([num, nums[left], nums[right]])
+
+                    # skip duplicated numbers ( ex. nums = [-3 0 0 0 3 3] )
+                    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 nums[left] + nums[right] < target:
+                    left += 1
+                else:
+                    right -= 1
+
+        return result
+    
+
+# Time Complexity: O(n^2)
+# - Sorting takes O(n log n).
+# - The outer loop runs O(n) times, and the two-pointer approach inside runs O(n) for each iteration.
+# - Combined, the overall time complexity is O(n^2).
+
+# Space Complexity: O(k)
+# - The result list uses O(k) space, where k is the number of unique triplets in the output.