Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of 

each bar is 1, compute how much water it is able to trap after raining.
For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Solution: Find left bound and right bound for each element. O(n).

class Solution {
public:
    int trap(int A[], int n) {
        int max_val = 0;
        int max_ind = 0;
        for(int i = 0; i < n; i++) {
            if(A[i] > max_val) {
                max_val = A[i];
                max_ind = i;
            }
        }
        
        int trap = 0;
        int start = 0;
        for(int i = start+1; i < max_ind; i++) {
            if(A[i] < A[start]) {
                trap += A[start]-A[i];
            }
            else {
                start = i;
            }
        }
        
        start = n-1; //update start
        for(int i = start-1; i > max_ind; i--) {
            if(A[i] < A[start]) {
                trap += A[start]-A[i];
            }
            else{
                start = i;
            }
        }
        return trap;
    }
};