42. Trapping Rain Water

Problem Statement

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

C++

Source file

class Solution {
public:
    int trap(vector<int>& ht) {
        int res=0, lt=0, rt=0, opq=0;
        for(int i=1; i<ht.size(); i++){
            if(ht[i]<ht[lt]){
                opq+=ht[i];
                rt=i;
            } else {
            res+=(rt-lt)*ht[lt]-opq;
            lt=rt=i;
            opq=0;
            }
            // cout<<i<<".\tLt:"<<lt<<"\tRt:"<<rt<<"\tOpq:"<<opq<<"\t"<<res<<endl;
        }
        int bound = lt;
        cout << bound;
        lt=ht.size()-1, rt=ht.size()-1, opq=0;
        for(int i=ht.size()-1; i>=bound; i--){
            if(ht[i]<ht[lt]){
                opq+=ht[i];
                rt=i;
            } else {
            res+=(lt-rt)*ht[lt]-opq;
            lt=rt=i;
            opq=0;
            }
            // cout<<i<<".\tLt:"<<lt<<"\tRt:"<<rt<<"\tOpq:"<<opq<<"\t"<<res<<endl;
        }
        return res;
    }
};