A solution and, importantly, a proof for LeetCode Problem 11 - Container with Most Water.

This problem intrigued me as, unlike many “LeetCode-esque” problems, little advanced or specific algorithms theory is required for an optimal $O(n)$ solution. That is, there was no algorithmic or mathematical *trick* that one needed to solve the problem. Instead, the solution relies on a subtle insight that leads to a relatively simple algorithm. This, in my opinion, is the real skill behind designing algorithms that can be gained from “LeetCode-esque” problems - as opposed to re-writing the same handful of algorithms over and over again.

The algorithm itself is fairly simple to understand intuitively, however understanding *why* it works is less so. In this article, I present my solution (which is essentially identical to many others you may find) and provide a **formal proof** of the key insight; something I have been unable to find a satisfactory example of.

I encourage the reader to visit LeetCode and attempt to solve the problem before reading further.

## Problem

Given $n$ non-negative integers $a_{1}, a_{2}, …, a_{n}$ , where each represents a point at coordinate $(i, a_{i})$. $n$ vertical lines are drawn such that the two endpoints of line $i$ is at $(i, a_{i})$ and $(i, 0)$. Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and $n \geq 2$

**Example**:

```
Input: [1, 8, 6, 2, 5, 4, 8, 3, 7]
Output: 49
```

## Solution

```
impl Solution {
pub fn max_area(height: Vec<i32>) -> i32 {
let mut l = 0;
let mut r = height.len() - 1;
let mut max_area = 0;
while l < r {
let width = (r - l) as i32;
if height[l] < height[r] {
max_area = max(max_area, height[l] * width);
l += 1;
} else {
max_area = max(max_area, height[r] * width);
r -= 1;
}
}
max_area
}
}
```

Using two pointers, the entire array of heights is processed in a single pass from *both ends* simultaneously; thus $O(n)$. On each iteration, the area of the container between `l`

and `r`

is calculated and the lower-heighted of the two is moved *towards* the other. Either can be moved in the case that `i == j`

(so long as it is consistent) - here I chose to move `r`

. The maximum are

The key idea is that for any two positions of `l`

and `r`

, the corresponding container area is *larger* than that of any container produced by moving the pointer with the larger height *towards* the other. **All such containers therefore need not be considered**.

## Proof

### Definitions

Let $a_{i}$ be the height of the line at index $i$ and $W(i, j)$, $H(i, j)$, $A(i, j)$ be the width, height and area, respectively, formed by the container between $a_{i}$ and $a_{j}$:

$$ W(i, j) = i-j $$

$$ H(i, j) = min(a_{i}, a_{j}) $$

$$ A(i, j) = W(i, j) \cdot H(i,j)$$

### Loop Invariant

The “key idea” previously mentioned.

*Hypothesis*: For any $i, j$ where $i< j$, $A(i, j)$ is greater than the area produced by moving either $j$ *towards* $i$ if $a_{i} < a_{j}$ or $i$ towards $j$ otherwise:

$$\begin{cases}max(\{A(i, j),…,A(i, i+1\}) &\text{if $a_{i} < a_{j}$}\\max(\{A(i, j),…,A(j-1, j\}) &\text{else}\end{cases}$$

When $a_{i} < a{j}$:

$$i^\prime=i$$

$$j^\prime = j - 1$$

The container width, by definition, has decreased since $j$ has moved closer to $i$:

$$W(i, j^\prime) < W(i, j)$$

Since $a_{i} < a_{j}$, there are two cases for the height of the container $H(i, j^\prime)$:

$$a_{i} \geq a_{j^\prime} \Rightarrow H(i, j^\prime) = a_{j^\prime} \leq H(i, j)$$

$$a_{i} < a_{j^\prime} \Rightarrow H(i, j^\prime) = a_{i} < H(i, j)$$

$$\Rightarrow H(i, j’) \leq H(i, j)$$

Thus, $$A(i, j^\prime) < A(i, j) \quad \text{by definition of \(A\)}$$

The same process applies to the case where $a_{i} > a_{j}$.

### Termination

Since the result is returned immediately after exiting the single `while`

loop, proving termination is equivalent to proving the termination of this loop.

The `while`

loop terminates when the expression `l < r`

evaluates to `false`

:

```
while l < r {
// ...
}
// l >= r
```

The loop contains and `if`

-`else`

expression which defined the *only two* possible code paths:

```
while l < r {
let width = (r - l) as i32;
if height[l] < height[r] {
max_area = max(max_area, height[l] * width);
l += 1; // 1.
} else {
max_area = max(max_area, height[r] * width);
r -= 1; // 2.
}
}
```

The lines indicated `1.`

and `2.`

are the only lines that affect the `while`

loop condition - calculations of `width`

and `max_area`

do not alter the values of `l`

and `r`

. Therefore, there are two relevant cases:

```
while l < r {
if some_condition {
l += 1;
} else {
r -= 1;
}
}
```

Either, `l`

is incremented or `r`

is decremented. Since `l`

is initialised to `0`

and `r`

to the length of the `height`

array the two **must converge** and thus the `while`

loop **must terminate**.

More formally, with respect to pointers $l$ and $r$, each iteration of the `while`

produces new values of $l$ and $r$ as follows:

$$(l^\prime, r^\prime) =\begin{cases}(l + 1, r) &\text{if $a_{i} < a_{j}$}\\(l, r - 1) &\text{else}\end{cases}$$

Since $l$ and $r$ are initialised to $0$ and $n$, where $n > 0$, the two values converge and thus loop exit condition $l < r$ is satisfied.