Submit Two-Pointers-1 PR

Problem_1.py

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+# https://leetcode.com/problems/sort-colors/
+
+# Time complexity: O(n) 
+# Space complexity: O(1)
+# Explanation: Use 3 pointers, and move every 0 to the left, every 2 to the right, and keep the 1s in the middle;
+# while swapping make sure left side swaps increment both p0 and curr pointer because those values have been checked
+# but right side swaps increment only p2 because those values have not been checked
+
+class Solution:
+    def sortColors(self, nums: List[int]) -> None:
+        """
+        Do not return anything, modify nums in-place instead.
+        """
+        n = len(nums)
+        p0, p2 = 0, n - 1 
+        curr = 0
+        while curr <= p2:
+            if nums[curr] == 0:
+                nums[p0], nums[curr] = nums[curr], nums[p0]
+                p0 += 1
+                curr += 1
+            elif nums[curr] == 2:
+                nums[p2], nums[curr] = nums[curr], nums[p2]
+                p2 -= 1
+            else:
+                curr += 1
+
+

Problem_2.py

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+# https://leetcode.com/problems/3sum/
+
+# Time complexity: O(n^2) 
+# Space complexity: O(n)
+# Explanation: Take each number that is less than 0, and try to find subarrays whhere sum is 0;
+# to find subarray assume the search space is for current index i to end of array
+
+class Solution:
+    def twoSum(self, nums: list[int], i: int, res: list[list[int]]):
+        low, high = i + 1, len(nums) - 1
+        while low < high:
+            sum = nums[i] + nums[low] + nums[high]
+            if sum < 0:
+                low += 1
+            elif sum > 0:
+                high -= 1
+            else:
+                res.append([nums[i], nums[low], nums[high]])
+                low += 1
+                high -= 1
+                while low < high and nums[low - 1] == nums[low]:
+                    low += 1
+
+
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        res = []
+        nums.sort()
+        for i in range(len(nums)):
+            if nums[i] > 0:
+                break
+            if i == 0 or nums[i - 1] != nums[i]:
+                self.twoSum(nums, i, res)
+        return res

Problem_3.py

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+# https://leetcode.com/problems/container-with-most-water/
+
+# TC: O(n)
+# SC: O(1)
+# Explanation: Maintain two pointers starting at index 0 and n - 1; keep moving the pointers based on highest point;
+# at each pass find the max area; return the final max area
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        n = len(height)
+        l, r = 0, n - 1
+        m_height = 0
+
+        while l < r:
+            m_height = max(m_height, min(height[l], height[r]) * (r - l))
+            if height[l] < height[r]:
+                l += 1
+            else:
+                r -= 1
+
+        return m_height