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In this post, you will find the solution for the **Container With Most Water** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Container With Most Water– LeetCode Problem

Container With Most Water– LeetCode Problem

**Problem:**

You are given an integer array `height`

of length `n`

. There are `n`

vertical lines are drawn such that the two endpoints of the ith line is `(i, 0)`

and `(i, height[i])`

.

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return *the maximum amount of water a container can store*.

**Notice** that you may not slant the container.

**Example 1:**

Input:height = [1,8,6,2,5,4,8,3,7]Output:49Explanation:The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

**Example 2:**

Input:height = [1,1]Output:1

**Constraints:**

`n == height.length`

`2 <= n <= 10`

^{5}`0 <= height[i] <= 10`

^{4}

Container With Most Water– LeetCode Solutions

class Solution { public: int maxArea(vector<int>& height) { int ans = 0; int l = 0; int r = height.size() - 1; while (l < r) { const int minHeight = min(height[l], height[r]); ans = max(ans, minHeight * (r - l)); if (height[l] < height[r]) ++l; else --r; } return ans; } };

class Solution { public int maxArea(int[] height) { int ans = 0; int l = 0; int r = height.length - 1; while (l < r) { final int minHeight = Math.min(height[l], height[r]); ans = Math.max(ans, minHeight * (r - l)); if (height[l] < height[r]) ++l; else --r; } return ans; } }

class Solution: def maxArea(self, height: List[int]) -> int: ans = 0 l = 0 r = len(height) - 1 while l < r: minHeight = min(height[l], height[r]) ans = max(ans, minHeight * (r - l)) if height[l] < height[r]: l += 1 else: r -= 1 return ans