PROBLEM LINK:
Div1, Div2
Practice
Author:Praveen Dhinwa
Tester:Triveni Mahatha
Editorialist:Adarsh Kumar
DIFFICULTY:
Easy-Medium
PREREQUISITES:
Binary search
PROBLEM:
You are given an array $A$ of $N$ integers. You need to find the smallest $K$ that satisfies this inequality; $\sum \limits_{i=1}^N \left \lceil \frac{A[i]}{K} \right \rceil \le H$, where $\left \lceil \right \rceil$ indicates the ceil function.
EXPLANATION:
According to the problem statement, Chef will need $\left \lceil \frac{A[i]}{K} \right \rceil$ hours to finish the $i^{th}$ pile. Hence, we need to find the smallest $K$ that satisfies this inequality; $\sum \limits_{i=1}^N \left \lceil \frac{A[i]}{K} \right \rceil \le H$.
subtask #1
Iterate over values of $K$ from $1$ to $MAX$ and break whenever you find the solution. Time complexity for the same will be $O(N.MAX)$, where $MAX$ is maximum element that can be present in the array.
subtask #2
For this subtask we need something better than brute-force. Hence, we will try to make some-observations first. Lets name the left side of our function as cost function. Observe that, our cost function is inversely proportional to $K$. Our cost function will decrease while $K$ increases. At some point it will become smaller than $H$. We need to find this point. Observe that, this problem has reduced to standard formulation of binary search problem. We just need to binary search on values of $K$ now and change the limits of $K$ according to the difference between our cost function and $H$. For more implementation details, you can have a look at attached solutions.
Time Complexity:
$O(N.log(MAX))$