Rainwater harvesting

#1

did not understand the solution

#2

@shikharskj The idea is to traverse every array element and find the highest bars on left and right sides. Take the smaller of two heights. The difference between the smaller height and height of the current element is the amount of water that can be stored in this array element.
To make this efficient one must pre-compute the highest bar on the left and right of every bar in linear time. Then use these pre-computed values to find the amount of water in every array element.
try to make code based on this logic and if you face any problem let me know.

#3

can u cALLME… 6394816023

#4

int maxWater(int arr[], int n)
{

// To store the maximum water  
// that can be stored 
int res = 0; 

// For every element of the array 
for (int i = 1; i < n-1; i++) { 

    // Find the maximum element on its left 
    int left = arr[i]; 
    for (int j=0; j<i; j++) 
       left = max(left, arr[j]); 

    // Find the maximum element on its right    
    int right = arr[i]; 
    for (int j=i+1; j<n; j++) 
       right = max(right, arr[j]);  

   // Update the maximum water     
   res = res + (min(left, right) - arr[i]);    
} 

return res;

}

#5

if you have any further queries let me know else please mark this doubt as resolved.

#6

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