Sync LeetCode submission Runtime - 565 ms (36.43%), Memory - 17.9 MB (73.40%)

Author: jimit105

0259-3sum-smaller/solution.py

@@ -1,28 +1,32 @@
-# Approach 3 - Two Pointers
+# Approach 3: Two Pointers
 
-# Time: O(n^2)
+# Time: O(n ^ 2)
 # Space: O(1)
 
 class Solution:
     def threeSumSmaller(self, nums: List[int], target: int) -> int:
+        if len(nums) < 3:
+            return 0
+
         nums.sort()
-        total = 0
-        
-        for i in range(len(nums)-1):
-            total += self.twoSumSmaller(nums, i + 1, target - nums[i])
-            
-        return total
-    
-    def twoSumSmaller(self, nums, start_index, target):
-        total = 0
-        left, right = start_index, len(nums) - 1
-        
-        while left < right:
-            if nums[left] + nums[right] < target:
-                total += right - left
-                left += 1
-            else:
-                right -= 1
-                
-        return total
-        
+        count = 0
+        n = len(nums)
+
+        # Fix the first element and use two pointers for the remaining elements
+        for i in range(n - 2):
+            left = i + 1
+            right = n - 1
+
+            while left < right:
+                curr_sum = nums[i] + nums[left] + nums[right]
+
+                if curr_sum < target:
+                    # All numbers between left and right will also work with left
+                    # because array is sorted and we want sum < target
+                    count += right - left
+                    left += 1
+                else:
+                    right -= 1
+
+        return count
+