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.

Python3JavaC++GoTypeScriptRustC#PHP

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))

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;
    }
}

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;
    }
};

func trap(height []int) (ans int) {
    n := len(height)
    left := make([]int, n)
    right := make([]int, n)
    left[0], right[n-1] = height[0], height[n-1]
    for 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])
    }
    for i, h := range height {
        ans += min(left[i], right[i]) - h
    }
    return
}

function trap(height: number[]): number {
    const n = height.length;
    const left: number[] = new Array(n).fill(height[0]);
    const right: number[] = new Array(n).fill(height[n - 1]);
    for (let 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]);
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += Math.min(left[i], right[i]) - height[i];
    }
    return ans;
}

impl Solution {
    #[allow(dead_code)]
    pub fn trap(height: Vec<i32>) -> i32 {
        let n = height.len();
        let mut left: Vec<i32> = vec![0; n];
        let mut right: Vec<i32> = vec![0; n];

        left[0] = height[0];
        right[n - 1] = height[n - 1];

        // Initialize the left & right vector
        for i in 1..n {
            left[i] = std::cmp::max(left[i - 1], height[i]);
            right[n - i - 1] = std::cmp::max(right[n - i], height[n - i - 1]);
        }

        let mut ans = 0;

        // Calculate the ans
        for i in 0..n {
            ans += std::cmp::min(left[i], right[i]) - height[i];
        }

        ans
    }
}

public 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;
    }
}

class Solution {
    /**
     * @param integer[] $height
     * @return integer
     */

    function trap($height) {
        $n = count($height);

        if ($n == 0) {
            return 0;
        }

        $left = 0;
        $right = $n - 1;
        $leftMax = 0;
        $rightMax = 0;
        $ans = 0;

        while ($left < $right) {
            if ($height[$left] < $height[$right]) {
                if ($height[$left] > $leftMax) {
                    $leftMax = $height[$left];
                } else {
                    $ans += $leftMax - $height[$left];
                }
                $left++;
            } else {
                if ($height[$right] > $rightMax) {
                    $rightMax = $height[$right];
                } else {
                    $ans += $rightMax - $height[$right];
                }
                $right--;
            }
        }
        return $ans;
    }
}

42. Trapping Rain Water

Description

Solutions

Solution 1: Dynamic Programming

Comments