LeetCode-in-All

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
• 1 <= n <= 2 * 104
• 0 <= height[i] <= 105

Solution

#include <stdio.h>

int trap(int* height, int heightSize) {
    int l = 0;
    int r = heightSize - 1;
    int res = 0;
    int lowerWall = 0;

    while (l < r) {
        int lVal = height[l];
        int rVal = height[r];

        // Determine which side has the lower wall and add water accordingly
        if (lVal < rVal) {
            // Update lower wall height if lVal has increased
            lowerWall = lVal > lowerWall ? lVal : lowerWall;
            // Add the trapped water at this position
            res += lowerWall - lVal;
            // Move left pointer
            l++;
        } else {
            // Update lower wall height if rVal has increased
            lowerWall = rVal > lowerWall ? rVal : lowerWall;
            // Add the trapped water at this position
            res += lowerWall - rVal;
            // Move right pointer
            r--;
        }
    }

    return res;
}

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