42. Trapping Rain Water

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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:

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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:

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

Constraints:

  • n == height.length
  • 1 <= n <= 2 * 10^4
  • 0 <= height[i] <= 10^5

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int trap(vector<int>& height) {
        int left = 0, right = height.size() - 1, max_left = 0, max_right = 0, res = 0;
        while (left <= right) {
             max_left = max(height[left], max_left);
             max_right = max(height[right], max_right);
             if (max_left < max_right) {
                res += max_left - height[left];
                left++;
             } else {
                res += max_right - height[right];
                right--;
             }
        }
        return res;
    }
};

monotonic stack

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class Solution {
public:
    int trap(vector<int>& height) {
        stack<int> stk;
        int n = height.size(), res = 0;
        for (int i = 0; i < n; i++) {
            int h = height[i];
            while (!stk.empty() && height[stk.top()] < h) {
                int bottom = height[stk.top()];
                stk.pop();
                if (stk.empty()) {
                    break;
                }
                int left = stk.top();
                res += (min(h, height[left]) - bottom) * (i - left - 1);
            }
            stk.push(i);
        }
        return res;
    }
};