712107. Koko Eating Bananas
Koko Eating Bananas
Slug: koko-eating-bananas
Difficulty: Medium
Id: 712107
Topic Tags: Binary Search, Arrays
Company Tags: Bloomberg, Amazon, Microsoft, Walmart, Adobe, Arcesium, Uber
Summary
Koko is eating bananas. She eats them in a specific order and we need to find the minimum number of operations required for her to finish eating all the bananas. This problem can be solved using binary search.
Detailed Explanation
The problem statement says that Koko has N bananas and she will eat them one by one from left to right. If she eats an even-indexed banana, she will skip the next one, but if she eats an odd-indexed banana, she will also eat the next one. We need to find the minimum number of operations required for her to finish eating all the bananas.
To solve this problem, we can use binary search. We first sort the array and then perform a binary search on it. The idea is that if Koko eats an even-indexed banana at the beginning, she will skip the next one. If she eats an odd-indexed banana, she will also eat the next one.
Here is a step-by-step breakdown of the solution:
- Initialize two pointers, low and high, to the start and end of the array.
- While low < high:
- Calculate mid as (low + high) / 2.
- If mid is even, calculate newLow as mid + 1.
- Otherwise, calculate newHigh as mid - 1.
- Return the minimum number of operations required for Koko to finish eating all the bananas.
The time complexity of this solution is O(logN) and the space complexity is O(1).
Optimized Solutions
Java
public class Solution {
public int kokoEatingBananas(int[] piles) {
Arrays.sort(piles);
int low = 0, high = piles.length;
while (low < high) {
int mid = (low + high) / 2;
if ((mid % 2 == 1 && piles[mid] <= piles[high - 1]) || (mid % 2 == 0 && piles[mid] >= piles[high - 1])) {
high = mid;
} else {
low = mid + 1;
}
}
return piles.length - low;
}
}