**source:**https://leetcode.com/problems/container-with-most-water**C/C++**

**Solution to LeetCode**problem

**11**.

**Container With Most Water**.

## Problem

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.

## Examples

**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

## Constraints

`n == height.length`

`2 <= n <= 10`

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

^{4}

## Solution

We will use two pointers (indexes), `l`

(left) and `r`

(right):

`l`

starts at index`0`

.`r`

starts at the last index.- While
`l < r`

:- Calculate area (
`shortest_bar * distance_between_bars`

) - If the area is bigger, then we update the area.
- Move the pointers:
- If
`l`

bar shorter, then move`l`

to the right. - Else move
`r`

to the left.

- If

- Calculate area (
- Return the max area.

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