## 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))

## AUTHORâ€™S AND TESTERâ€™S SOLUTIONS

Setterâ€™s solution

Testerâ€™s solution

2 Likes

why canâ€™t I see the solution?? the links donâ€™t work

unable to see the solutions , bucket policy needs to be updated .

I have a doubt.

For the following test case:

```
5 7
9 3 8 10 5
```

The answer should be 6. So taking k=6, the above inequality will be calculated as following:

```
9/6 + 3/6 + 8/6 + 10/6 + 5/6 (Ceil of all the terms are added)
2 + 1 + 2 + 2 + 1
=8
```

Hour available is 7, so am I doing something. If thereâ€™s any problem in logic, please point out the correct solution to above test case and logic.

Thanks

@philomath_mht the ans. will be 8 as you can only divide maximum (7-5)=2 bunch of bananas and the minimum required bananas that should be eaten is 8.

9/8 + 3/8 + 8/8 + 10/8 + 5/8 = 2 + 1 + 1 + 2 + 1 = 7

Change int to long long && float to double. Hereâ€™s your modified solution :

https://www.codechef.com/viewsolution/17827412

@dharmick

! Should have tried this. Was just too lazy to do it

Please tell whatâ€™s wrong with my solution

here

you can look at my code. its almost similar to your code(99%same)

link -**https://www.codechef.com/viewsolution/17841451**

Do upvote if you find useful.

https://www.codechef.com/viewsolution/17841507

Can someone tell me whatâ€™s wrong with this???

https://www.codechef.com/viewsolution/17889934

can someone tell me whatâ€™s wrong with this??

Can someone tell me what is the mistake in my code:

https://www.codechef.com/viewsolution/17768075

I was getting a PA with my solution and just after I changed

```
**
ans += std::ceil((float)a[i]/k); to
ans += (a[i] + hr - 1)/hr ;
**
```

it go ACâ€¦**will someone please explain how the latter is different then the first one?**