Trapping Rain Water

/*
Basic logic:we will find water trapped onto every building by min(left_max,right_max)-arr[i];

T.C : O(n),S.C O(1),one pass
In our algo,left_max will store maximum of 0 to i-1 and right_max will store maximum of j+1 to n-1.
using two pointer(i,j),so if we think about ith index,so we have left_max for 0 to i-1 and right_max from j+1 to n-1
so,
Case 1: if left_max < right_max:
 *We will think to get water over the ith building.
 we need max of 0 to i-1 that we already have ie. left_max
 we need max of i+1 to n-1 but we have maxiumum of j+1 to n-1 only,not having maxiumum of i+1 to j
 but even if some building height is greater than right_max still we will have min(left_max,right_max) = left_max
 because left_max is smaller than even current right_max.
Case 2: if left_max>=right_max:
 *We will think to get water over jth building.
 If left_max is greater and right_max is smaller then we need only smaller one so we got the water height.
*/

class Solution{

    // Function to find the trapped water between the blocks.
    public:
    long long trappingWater(int arr[], int n){
        long long int left_max=arr[0],right_max=arr[n-1];
        long long int i=1,j=n-2,ans=0;
        while(i<=j){
           if(left_max<right_max){
               if(arr[i]>left_max) left_max=arr[i];  //it's top can't hold water because left side has no boundary
               else{
                   ans+=(left_max-arr[i]);
               }
               i++;
           }
           else{
               if(arr[j]>right_max)  right_max=arr[j];  //it's top can't hold because right side has no boundary
               else{
                   ans+=(right_max-arr[j]);
               }
               j--;
           }
        }
        return ans;
    }
};