add solution: 3sum

Author: dusunax

3sum/dusunax.py

@@ -0,0 +1,50 @@
+'''
+# Leetcode 15. 3Sum
+
+use **two pointers** to solve this problem.
+
+## Time and Space Complexity
+
+```
+TC: O(n^2)
+SC: O(1)
+```
+
+### TC is O(n^2):
+- sorting the list = O(n log n)
+- iterating through the list and using two pointers to find the sum of three numbers. = O(n^2)
+
+### SC is O(1):
+- sorting in place = O(1)
+'''
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        nums.sort() # TC: O(n log n), SC: O(1)
+        result = [] # result are part of the output => do not count toward auxiliary (extra) space.
+        
+        for i in range(len(nums)): # TC: O(n^2)
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+                
+            j = i + 1
+            k = len(nums) - 1
+            while j < k:
+                currSum = nums[i] + nums[j] + nums[k]
+
+                if currSum < 0:
+                    j += 1
+                elif currSum > 0:
+                    k -= 1
+                else:
+                    result.append([nums[i], nums[j], nums[k]]) 
+
+                    while j < k and nums[j] == nums[j + 1]:
+                        j += 1
+                    while j < k and nums[k] == nums[k - 1]: 
+                        k -= 1
+
+                    j += 1
+                    k -= 1
+        
+        return result