Merge pull request #7 from xchux/11-container-with-most-water

✨ (solution): Problem 11. Container With Most Water

Author: xchux

solutions/11. Container With Most Water/README.md

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+---
+comments: true
+difficulty: easy
+# Follow `Topics` tags
+tags:
+    - Array
+    - Two Pointers
+    - Greedy
+---
+
+# [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/description/)
+
+## Description
+
+Given an integer array `height` of length `n`, where each element represents the height of a vertical line drawn at position `i`, with endpoints at `(i, 0)` and `(i, height[i])`, your task is to determine the maximum amount of water that can be contained between any two lines and the x-axis.
+
+Return the maximum water volume that the container can store.
+
+**Example 1:**
+```
+Input: height = [1,6,5,5,4,1,7,1]
+Output: 36
+Explanation: In this case, the max area of water ([6,5,5,4,1,7]) the container can contain is 36.
+```
+
+**Example 2:**
+```
+Input: nums = [2,2]
+Output: 2
+```
+
+
+**Constraints:**
+`n == height.length`
+`2 <= n <= 10^5`
+`0 <= height[i] <= 10^4`
+
+## Solution
+
+Simply put, we calculate the maximum area of ​​the rectangle and return it. The formula shrinks inward, with the highest point as the side.
+
+```java
+class Solution {
+    public int maxArea(int[] height) {
+        int left = 0, right = height.length - 1;
+        int maxArea = 0;
+        while (left < right) {
+            int area = (right - left) * Math.min(height[left], height[right]);
+            maxArea = Math.max(maxArea, area);
+            if (height[left] < height[right]) {
+                left++;
+            } else {
+                right--;
+            }
+        }
+        return maxArea;
+    }
+}
+```
+
+```python
+class Solution:
+    def maxArea(self, height: list[int]) -> int:
+        left, right = 0, len(height) - 1
+        max_area = 0
+        while left < right:
+            max_area = max(max_area, min(height[left], height[right]) * (right - left))
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+        return max_area
+```
+
+
+## Complexity
+
+- Time complexity: $$O(n^2)$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->
+

solutions/11. Container With Most Water/Solution.java

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+class Solution {
+    public int maxArea(int[] height) {
+        int left = 0, right = height.length - 1;
+        int maxArea = 0;
+        while (left < right) {
+            int area = (right - left) * Math.min(height[left], height[right]);
+            maxArea = Math.max(maxArea, area);
+            if (height[left] < height[right]) {
+                left++;
+            } else {
+                right--;
+            }
+        }
+        return maxArea;
+    }
+}

solutions/11. Container With Most Water/Solution.py

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+class Solution:
+    def maxArea(self, height: list[int]) -> int:
+        left, right = 0, len(height) - 1
+        max_area = 0
+        while left < right:
+            max_area = max(max_area, min(height[left], height[right]) * (right - left))
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+        return max_area