Water Supply Complete Solution

This article is part of the Scaler Topics Fortnightly Contest - 16

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Solution Approach

Closing outgoing hole of large taps is similar to selecting taps with smaller outgoing hole and closing rest of them.
So, first we can reverse the array C. So, that the last tap which should always be open come to first position.
Now, we can sort the array from second element to last. So, that we can take them one by one and closing rest of all. Now, our equation is (C[i]*A)/X, where X is the sum of hole size of all the open taps. So, now last tap has become first one. So, (C[0]*A)/X >= B.
Now, it becomes $C[0]*A$ >= X*B
This means we can iterate on the sorted array C and take all taps one by one until we get $C[0]*A$ < X*B, where X is the sum of taps taken till now. And if all taps are taken than answer is 0.