README.md

42. Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.

Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9

Constraints:

• n == height.length
• 1 <= n <= 2 * 104
• 0 <= height[i] <= 105

Solutions (Rust)

1. Solution

impl Solution {
    pub fn trap(height: Vec<i32>) -> i32 {
        let mut left_max = vec![0; height.len()];
        let mut right_max = vec![0; height.len()];

        for i in 1..height.len() {
            left_max[i] = left_max[i - 1].max(height[i - 1]);
        }
        for i in (0..height.len() - 1).rev() {
            right_max[i] = right_max[i + 1].max(height[i + 1]);
        }

        (0..height.len())
            .map(|i| (left_max[i].min(right_max[i]) - height[i]).max(0))
            .sum()
    }
}