# 11. Container with Most Water
> [!META]- Inline Metadata
> [tags:: #career #concepts/programming/algorithms/two-pointers ]
> [project group:: Neetcode 150]
> [difficulty:: medium]
> [status:: done]
> [link:: https://leetcode.com/problems/container-with-most-water/]
> [up:: [[Leetcode MOC]]]
## Problem
You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(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
### Constraints:
- `n == height.length`
- `2 <= n <= 105`
- `0 <= height[i] <= 104`
## Solution
### Mine
Start with brute force: two pointers, go through every combo of left and right indices and keep the biggest area. $O(n^2)$
Improvements:
- Start with right pointer at end of array
- Change rule for shifting pointers: shift only the pointer from the smallest height.
- If heights are equal, shift either one
- When pointers are equal, end
#### Python
```python
def most_water(container: List[int]) -> int:
max_area = 0
i = 0
j = len(container) - 1
while i != j:
max_area = max(max_area, min(container[i], container[j]) * (j - i))
if container[i] < container[j]:
i += 1
else:
j -= 1
return max_area
```
#### Rust
### LC's
## Things I Learned
## Scratchpad