Merge pull request neetcode-gh#2505 from sgrtye/revert-wrong-commit

[Bugfix][Python][0015-3sum] Revert wrong commit

Author: neetcode-gh

python/0015-3sum.py

@@ -1,27 +1,28 @@
 class Solution:
-    def ThreeSum(self, integers):
-        """
-        :type integers: List[int]
-        :rtype: List[List[int]]
-        """
-        integers.sort()
-        result = []
-        for index in range(len(integers)):
-            if integers[index] > 0:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        res = []
+        nums.sort()
+
+        for i, a in enumerate(nums):
+            # Skip positive integers
+            if a > 0:
                 break
-            if index > 0 and integers[index] == integers[index - 1]:
+
+            if i > 0 and a == nums[i - 1]:
                 continue
-            left, right = index + 1, len(integers) - 1
-            while left < right:
-                if integers[left] + integers[right] < 0 - integers[index]:
-                    left += 1
-                elif integers[left] + integers[right] > 0 - integers[index]:
-                    right -= 1
+
+            l, r = i + 1, len(nums) - 1
+            while l < r:
+                threeSum = a + nums[l] + nums[r]
+                if threeSum > 0:
+                    r -= 1
+                elif threeSum < 0:
+                    l += 1
                 else:
-                    result.append([integers[index], integers[left], integers[right]]) # After a triplet is appended, we try our best to incease the numeric value of its first element or that of its second.
-                    left += 1 # The other pairs and the one we were just looking at are either duplicates or smaller than the target.
-                    right -= 1 # The other pairs are either duplicates or greater than the target.
-                    # We must move on if there is less than or equal to one integer in between the two integers.
-                    while integers[left] == integers[left - 1] and left < right:
-                        left += 1 # The pairs are either duplicates or smaller than the target.
-        return result
+                    res.append([a, nums[l], nums[r]])
+                    l += 1
+                    r -= 1
+                    while nums[l] == nums[l - 1] and l < r:
+                        l += 1
+                        
+        return res