42-trapping-rain-water.cpp

/*
42. Trapping Rain Water
Hard - 66.1%

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
*/

#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int trap(vector<int>& height) {
        if (height.empty()) return 0;
        
        int left = 0, right = height.size() - 1;
        int leftMax = 0, rightMax = 0;
        int water = 0;
        
        while (left < right) {
            if (height[left] < height[right]) {
                // Process left side
                if (height[left] >= leftMax) {
                    leftMax = height[left];
                } else {
                    water += leftMax - height[left];
                }
                left++;
            } else {
                // Process right side
                if (height[right] >= rightMax) {
                    rightMax = height[right];
                } else {
                    water += rightMax - height[right];
                }
                right--;
            }
        }
        
        return water;
    }
};