Trapping Rain water (#42 Leetcode)

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

1. Increment the left position until we find the left element greater than right. 

2. Water can only be trapped in the heigh[i] if the maxHeight is greater than height[i].

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

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