# Eat Bananas in K Hours - Facebook Top Interview Questions

### Problem Statement :

```You are given a list of integers piles and an integer k.

piles[i] represents the number of bananas on pile i. On each hour, you can choose any pile and eat r number of bananas in that pile.

If you pick a pile with fewer than r bananas, it still takes an hour to eat the pile.

Return the minimum r required such that you can eat all the bananas in less than or equal to k hours.

Constraints

n ≤ 100,000 where n is the length of piles

n ≤ k

Example 1

Input

piles = [6, 4, 3]

k = 5

Output

3

Explanation

At r = 3 bananas per hour, we can eat the first pile in 2 hours, eat the second in 2 hours, and eat the last
pile in 1 hour.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<int>& piles, int k) {
int s = 1, e = *max_element(piles.begin(), piles.end());
auto check = [&piles, &k](int r) {
int count = 0;
for (int p : piles) {
count += (p + r - 1) / r;
}
return count <= k;
};

while (s <= e) {
int m = s + (e - s) / 2;
if (check(m)) {
e = m - 1;
} else {
s = m + 1;
}
}
return s;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] A, int k) {
int lo = 1;
int hi = Integer.MAX_VALUE;
int r = -1;

while (lo <= hi) {
int mid = lo + (hi - lo) / 2; // speed

if (timeToEat(A, mid) <= k) { // valid, try to eat slower, go left.
hi = mid - 1;
r = mid;
} else { // couldn't eat, need to eat faster, go right.
lo = mid + 1;
}
}

return r;
}

public int timeToEat(int[] A, int speed) {
int r = 0;

for (int x : A) {
r += x / speed;
if (x % speed != 0)
r++;
}

return r;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, piles, k):
lo = 1
hi = max(piles)

while lo <= hi:
r = (lo + hi) // 2

if self.bananas_hour(piles, r) <= k:
hi = r - 1
else:
lo = r + 1
return lo

def bananas_hour(self, piles, max_bananas_hour):
count_hours = 0

for pile in piles:
count_hours += pile // max_bananas_hour

if pile % max_bananas_hour != 0:
count_hours += 1

return count_hours```
```

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