Last updated on October 5th, 2024 at 05:49 pm

Here, We see ** Container With Most Water LeetCode Solution**. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Array, Two-Pointers

## Companies

Bloomberg

## Level of Question

Medium

**Container With Most Water LeetCode Solution**

## Table of Contents

**Problem Statement**

You are given an integer array `height`

of length `n`

. There are `n`

vertical lines drawn such that the two endpoints of the `i`

line are ^{th}`(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:** 49 **Explanation:** 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

**1. Container With Most Water LeetCode Solution C++**

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

**2. Container With Most Water Solution Java**

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

**3. Container With Most Water Solution JavaScript**

var maxArea = function(height) { let ans = 0, i = 0, j = height.length-1 while (i < j) { ans = Math.max(ans, Math.min(height[i], height[j]) * (j - i)) height[i] <= height[j] ? i++ : j-- } return ans };

**4. Container With Most Water Solution Python**

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