description

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

submission

// see: https://leetcode.com/problems/trapping-rain-water/solutions/6563751/an-optical-approach-to-the-water-trapping-problem
impl Solution {
    pub fn trap(height: Vec<i32>) -> i32 {
        let f_fold = |(a, mut m): (i32, i32), val: &i32| {
            m = m.max(*val);
            (a + m, m)
        };
        let (a_l, max_height) = height.iter().fold((0, 0), f_fold);
        let (a_r, _) = height.iter().rev().fold((0, 0), f_fold);
        let a_box = max_height * height.len() as i32;
        let a_terrain: i32 = height.iter().sum();
        (a_l + a_r) - (a_box + a_terrain)
    }
}