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42. Trapping Rain Water

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

Solutions

Solution 1: Dynamic Programming

We define $left[i]$ as the height of the highest bar to the left of and including the position at index $i$, and $right[i]$ as the height of the highest bar to the right of and including the position at index $i$. Therefore, the amount of rainwater that can be trapped at index $i$ is $min(left[i], right[i]) - height[i]$. We traverse the array to calculate $left[i]$ and $right[i]$, and the final answer is $\sum_{i=0}^{n-1} \min(left[i], right[i]) - height[i]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

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class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        left = [height[0]] * n
        right = [height[-1]] * n
        for i in range(1, n):
            left[i] = max(left[i - 1], height[i])
            right[n - i - 1] = max(right[n - i], height[n - i - 1])
        return sum(min(l, r) - h for l, r, h in zip(left, right, height))

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class Solution {
    public int trap(int[] height) {
        int n = height.length;
        int[] left = new int[n];
        int[] right = new int[n];
        left[0] = height[0];
        right[n - 1] = height[n - 1];
        for (int i = 1; i < n; ++i) {
            left[i] = Math.max(left[i - 1], height[i]);
            right[n - i - 1] = Math.max(right[n - i], height[n - i - 1]);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += Math.min(left[i], right[i]) - height[i];
        }
        return ans;
    }
}

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class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        int left[n], right[n];
        left[0] = height[0];
        right[n - 1] = height[n - 1];
        for (int i = 1; i < n; ++i) {
            left[i] = max(left[i - 1], height[i]);
            right[n - i - 1] = max(right[n - i], height[n - i - 1]);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += min(left[i], right[i]) - height[i];
        }
        return ans;
    }
};

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func trap(height []int) (ans int) {
    n := len(height)
    left := make([]int, n)
    right := make([]int, n)
    left[0], right[n-1] =