Trapping Rain Water.cpp

/*
Trapping Rain Water
===================

Given an array arr[] of N non-negative integers representing the height of blocks. If width of each block is 1, compute how much water can be trapped between the blocks during the rainy season. 

Example 1:
Input:
N = 6
arr[] = {3,0,0,2,0,4}
Output:
10

Example 2:
Input:
N = 4
arr[] = {7,4,0,9}
Output:
10
Explanation:
Water trapped by above 
block of height 4 is 3 units and above 
block of height 0 is 7 units. So, the 
total unit of water trapped is 10 units.

Example 3:
Input:
N = 3
arr[] = {6,9,9}
Output:
0
Explanation:
No water will be trapped.

Your Task:
You don'y need to read input or print anything. The task is to complete the function trappingWater() which takes arr and N as input parameters and returns the total amount of water that can be trapped.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)

Constraints:
3 <= N <= 107
0 <= Ai <= 108
*/

// function to find the trapped water in between buildings
// arr: input array
// n: size of array
int trappingWater(int arr[], int n)
{
  int ans = 0;
  vector<int> left(n, 0), right(n, 0);
  left[0] = arr[0], right[n - 1] = arr[n - 1];

  for (int i = 1; i < n; ++i)
    left[i] = max(left[i - 1], arr[i]);
  for (int i = n - 2; i >= 0; --i)
    right[i] = max(right[i + 1], arr[i]);

  for (int i = 0; i < n; ++i)
  {
    int level = min(left[i], right[i]);
    if (level > arr[i])
      ans += level - arr[i];
  }
  return ans;
}