**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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