Completed TwoPointers-1

3Sum.py

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+# Time Complexity : O(n^2)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Sort the array to use two pointers.
+# Fix a number and look for two other numbers using left and righ pointers.
+# Skip duplicates to avoid repeating the same triplet in the result.
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        n = len(nums)
+        res = []
+
+        nums.sort()
+
+        for i in range(n):
+            if i != 0 and nums[i] == nums[i-1]:
+                continue
+
+            if nums[i] > 0:
+                break
+
+            left, right = i+1, n-1
+
+            while left < right:
+
+                total = nums[i] + nums[left] + nums[right]
+
+                if total == 0:
+                    res.append([nums[i], nums[left], nums[right]])
+                    left += 1
+                    right -= 1
+                    while left < right and nums[left] == nums[left - 1]:
+                        left += 1
+                    while left < right and nums[right] == nums[right + 1]:
+                        right -= 1
+                elif total > 0:
+                    right -= 1
+                else:
+                    left += 1
+
+        return res
+
+

ContainerWithMostWater.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Two pointers at both ends and keep checking the area between them.
+# Move the pointer of shorter line
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        maxAmt = 0
+        left, right = 0, len(height) - 1
+
+        while left < right:
+            w = right - left
+
+            if height[left] < height[right]:
+                h = height[left]
+                left += 1
+            else:
+                h = height[right]
+                right -= 1
+
+            currAmt = h * w
+            maxAmt = max(maxAmt, currAmt)
+
+        return maxAmt
+

SortColors.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : 3 pointers ans swap elements accordingly to have 0s then 1s and then 2s.
+# Move mid only if num is 0 or 1
+
+class Solution:
+    def sortColors(self, nums: List[int]) -> None:
+        """
+        Do not return anything, modify nums in-place instead.
+        """
+        left, mid, right = 0, 0, len(nums) - 1
+
+        while mid <= right:
+            if nums[mid] == 2:
+                nums[mid], nums[right] = nums[right], nums[mid]
+                right -= 1
+            elif nums[mid] == 0:
+                nums[left], nums[mid] = nums[mid], nums[left]
+                left += 1
+                mid += 1
+            else:
+                mid += 1
+