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container_with_most_water.rs
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#![allow(dead_code)]
/*
#11 Container With Most Water
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.
ex.
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.
Constraints:
n == height.length
2 <= n <= 10^5
0 <= height[i] <= 10^4
*/
/*
Naive approach
T: O(N^2)
S: O(1)
*/
pub fn solution_a(heights: Vec<u32>) -> u32 {
let mut max_area = 0;
'left: for (i, a) in heights.iter().enumerate() {
for (j, b) in heights.iter().enumerate().rev() {
if i == j {
continue 'left;
}
// constraint enforces that 2^32 - 1 > usize (10^5)
let i = i as u32;
let j = j as u32;
let area = (j - i) * a.min(b);
max_area = max_area.max(area);
}
}
max_area
}
/*
Sliding window
T: O(N)
S: O(1)
*/
pub fn solution_b(heights: Vec<u32>) -> u32 {
let mut max_area = 0;
let mut heights = heights.iter().enumerate();
let mut left = heights.next();
let mut right = heights.next_back();
while let (Some((i, a)), Some((j, b))) = (left, right) {
max_area = max_area.max(a.min(b) * (j - i) as u32);
if a <= b {
left = heights.next();
} else {
right = heights.next_back();
}
}
max_area
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn a_0() {
let input = vec![1, 8, 6, 2, 5, 4, 8, 3, 7];
assert_eq!(solution_a(input), 49);
}
#[test]
fn a_1() {
let input = vec![1, 1];
assert_eq!(solution_a(input), 1);
}
#[test]
fn b_0() {
let input = vec![1, 8, 6, 2, 5, 4, 8, 3, 7];
assert_eq!(solution_b(input), 49);
}
#[test]
fn b_1() {
let input = vec![1, 1];
assert_eq!(solution_b(input), 1);
}
}