Scanning from left to right and right to left.

Find the max height of left and right. The minimum one will be most possible height for current state that can bound the water. But one thing need to check:

whether current height is higher than the value.

1 class Solution { 2 public: 3     int trap(int A[], int n) { 4         if (n < 3) return 0; 5         vector
     
       left(n), right(n); 6         int result = 0, rec = A[0]; 7         for (int i = 1; i < n; i++) { 8             left[i] = max(rec, left[i-1]); 9             rec = max(rec, A[i]);10         }11         rec = A[n-1];12         for (int i = n-2; i >= 0; i--) {13             right[i] = max(rec, right[i+1]);14             rec = max(rec, A[i]);15         }16         for (int i = 0; i < n; i++) {17             result += (min(left[i], right[i]) - A[i]) > 0 ? (min(left[i], right[i]) - A[i]) : 0;18         }19         return result;20     }21 };