2 pointers done

Problem1_SortColors.py

@@ -0,0 +1,32 @@
+# Time Complexity : O(n) where n is the length of the array
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Approach: Use Dutch National Flag algorithm with three pointers (left, current, right).
+# Left pointer tracks boundary of 0s, right pointer tracks boundary of 2s, current pointer traverses.
+# Swap elements based on current value: 0 goes to left, 2 goes to right, 1 stays in middle.
+
+class Solution:
+    def sortColors(self, nums: List[int]) -> None:
+        """
+        Do not return anything, modify nums in-place instead.
+        """
+        left = 0
+        current = 0
+        right = len(nums) - 1
+
+        while current <= right:
+            if nums[current] == 0:
+                # Swap with left pointer and move both forward
+                nums[left], nums[current] = nums[current], nums[left]
+                left += 1
+                current += 1
+            elif nums[current] == 2:
+                # Swap with right pointer, only move right backward
+                nums[current], nums[right] = nums[right], nums[current]
+                right -= 1
+            else:
+                # nums[current] == 1, just move current forward
+                current += 1
+

Problem2_3Sum.py

@@ -0,0 +1,45 @@
+# Time Complexity : O(n^2) where n is the length of the array
+# Space Complexity : O(1) excluding the output array
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Approach: Sort the array first, then for each element, use two pointers to find pairs that sum to negative of current element.
+# Skip duplicates at each level (outer loop, left pointer, right pointer) to avoid duplicate triplets.
+# Move pointers based on sum comparison: if sum < 0, move left forward; if sum > 0, move right backward.
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        nums.sort()
+        result = []
+        n = len(nums)
+
+        for i in range(n - 2):
+            # Skip duplicates for the first element
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+
+            left = i + 1
+            right = n - 1
+
+            while left < right:
+                current_sum = nums[i] + nums[left] + nums[right]
+
+                if current_sum == 0:
+                    result.append([nums[i], nums[left], nums[right]])
+
+                    # Skip duplicates for left pointer
+                    while left < right and nums[left] == nums[left + 1]:
+                        left += 1
+                    # Skip duplicates for right pointer
+                    while left < right and nums[right] == nums[right - 1]:
+                        right -= 1
+
+                    left += 1
+                    right -= 1
+                elif current_sum < 0:
+                    left += 1
+                else:
+                    right -= 1
+
+        return result
+

Problem3_ContainerWithMostWater.py

@@ -0,0 +1,29 @@
+# Time Complexity : O(n) where n is the length of the array
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Approach: Use two pointers at both ends of the array, calculate area using shorter height and distance between pointers.
+# Move the pointer with smaller height inward, as moving the larger height pointer cannot increase the area.
+# Track maximum area encountered during traversal.
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        left = 0
+        right = len(height) - 1
+        max_area = 0
+
+        while left < right:
+            # Calculate current area
+            width = right - left
+            current_area = min(height[left], height[right]) * width
+            max_area = max(max_area, current_area)
+
+            # Move pointer with smaller height
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+
+        return max_area
+