875. Koko Eating Bananas

Prompt

Koko loves to eat bananas. There are n piles of bananas, the ith pile has piles[i] bananas. The guards have gone and will come back in h hours.

Koko can decide her bananas-per-hour eating speed of k. Each hour, she chooses some pile of bananas and eats k bananas from that pile. If the pile has less than k bananas, she eats all of them instead and will not eat any more bananas during this hour.

Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.

Return the minimum integer k such that she can eat all the bananas within h hours.

Constraints:

• 1 <= piles.length <= 104
• piles.length <= h <= 109
• 1 <= piles[i] <= 109

Examples

• Example 1:
  • Input: piles = [3,6,7,11], h = 8
  • Output: 4
• Example 2:
  • Input: piles = [30,11,23,4,20], h = 5
  • Output: 30
• Example 3:
  • Input: piles = [30,11,23,4,20], h = 6
  • Output: 23

Solutions

Binary Search Solution

In C++

constexpr inline int ceil_divide(int a, int b) {return (a + b - 1) / b;}
int minEatingSpeed(const vector<int>& piles, int hours) {
	size_t l = 1; 
	size_t r = *ranges::max_element(piles.begin(), piles.end());
	size_t min_k = r;
	
	while (l <= r) {
		size_t k = (r - l) / 2 + l;
		long hours_balance = (long)hours;
		for (int pile : piles) hours_balance -= ceil_divide(pile, k);
		if (hours_balance >= 0) {
			min_k = k;
			r = k - 1;
		} else {
			l = k + 1;
		}
	}
	return min_k;
}

Explanation

The idea is that the solution space is $[1,\max \text{piles}]$, so you can just binary search that space to find the answer. The heuristic is how many hours it would take to finish all the bananas, subtracted from the hours alloted. See Integer Ceil Divide.

Big O Analysis

• Let $n$ be the number of piles
• Let $m$ be the size of the largest pile

Time Complexity

$O(n\log m)$

Auxiliary Space Complexity

$O(1)$