42. Trapping Rain Water

42. Trapping Rain Water (Hard)

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
• 0 <= n <= 3 * 104
• 0 <= height[i] <= 105

Related Topics:
Array, Two Pointers, Dynamic Programming, Stack

Similar Questions:

• Container With Most Water (Medium)
• Product of Array Except Self (Medium)
• Trapping Rain Water II (Hard)
• Pour Water (Medium)

Solution 1.

The water that can be held at position i is max(0, min(left[i], right[i]) - A[i]) where left[i] is the maximum height to the left of i and right[i] is the maximum height to the right of i.

// OJ: https://leetcode.com/problems/trapping-rain-water/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int trap(vector<int>& A) {
        int N = A.size(), ans = 0;
        vector<int> left(N, 0), right(N, 0);
        for (int i = 1; i < N; ++i) left[i] = max(left[i - 1], A[i - 1]);
        for (int i = N - 2; i >= 0; --i) right[i] = max(right[i + 1], A[i + 1]);
        for (int i = 1; i < N - 1; ++i) ans += max(0, min(left[i], right[i]) - A[i]);
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/trapping-rain-water/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int trap(vector<int>& A) {
        int N = A.size(), ans = 0, mx = 0, left = 0;
        vector<int> right(N);
        for (int i = N - 2; i >= 0; --i) right[i] = max(right[i + 1], A[i + 1]);
        for (int i = 0; i < N; ++i) {
            ans += max(0, min(left, right[i]) - A[i]);
            left = max(left, A[i]);
        }
        return ans;
    }
};

Or use stack.

// OJ: https://leetcode.com/problems/trapping-rain-water/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int trap(vector<int>& A) {
        stack<int> right;
        int N = A.size(), ans = 0, left = 0;
        for (int i = N - 1; i >= 0; --i) {
            if (right.empty() || A[i] > A[right.top()]) right.push(i);
        }
        for (int i = 0; i < N; ++i) {
            if (right.top() == i) right.pop();
            int h = min(left, right.size() ? A[right.top()] : 0);
            ans += max(h - A[i], 0);
            left = max(left, A[i]);
        }
        return ans;
    }
};

Solution 2. Two Pointers

// OJ: https://leetcode.com/problems/trapping-rain-water/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int trap(vector<int>& A) {
        if (A.empty()) return 0;
        int N = A.size(), ans = 0, i = 0, j = N - 1, left = 0, right = 0;
        while (i < j) {
            if (A[i] < A[j]) {
                left = max(left, A[i]);
                ans += left - A[i++];
            } else {
                right = max(right, A[j]);
                ans += right - A[j--];
            }
        }
        return ans;
    }
};