Koko Eating Bananas in Java: Explanation & Practice

Minimum eating speed to finish in h hours

Problem summary

Koko can decide her eating speed k (bananas per hour). Each hour she eats k bananas from one pile. Find minimum k to finish all piles in h hours.

Starter code

public class Main {
    public static void main(String[] args) {
        int[] piles = {3, 6, 7, 11};
        int h = 8;
        
        // Binary search on eating speed k
        // Min k = 1, Max k = max pile size
        // For each k, calculate hours needed
        // Print: Min speed: 4
    }
}

Expected output and test cases

[3,6,7,11], h=8
```
Min speed: 4
```
[30,11,23,4,20], h=5
```
Min speed: 30
```
[30,11,23,4,20], h=6
```
Min speed: 23
```

Hints

Binary search on the answer (eating speed)
For speed k, hours needed = sum of ceil(pile[i]/k) for all piles
If hours needed <= h, try smaller speed (right = mid)
If hours needed > h, need faster speed (left = mid + 1)

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