DSA interview question

Koko Eating Bananas

Medium · Algorithms · Topics: binary search on answer, array

The problem

Koko has n piles of bananas and h hours before the guards return. Each hour she picks one pile and eats up to k bananas from it; if the pile has fewer than k, she finishes it and waits for the next hour. She wants to eat slowly, so she needs the minimum speed k that still finishes every pile within h hours. Trying every speed from 1 upwards is far too slow when piles hold 10⁹ bananas. But feasibility is monotonic: any speed faster than a working speed also works. So binary search on k.

Example: piles = [3, 6, 7, 11], h = 8 → 4.

Function signature

int minEatingSpeed(const vector<int>& piles, long long h);  // h >= number of piles

Test cases

Input `(piles, h)`	Output	Explanation
`([3, 6, 7, 11], 8)`	4	1 + 2 + 2 + 3 = 8 hours
`([30, 11, 23, 4, 20], 5)`	30	One hour per pile: the speed must equal the largest pile
`([30, 11, 23, 4, 20], 6)`	23
`([1000000000], 2)`	500000000
`([312884470], 312884469)`	2	At speed 1 she'd need one hour too many

(Confirm: h ≥ number of piles, so an answer always exists (speed = max pile); hours per pile = ceil(pile / k).)

How interviewers phrase it

"Koko eating bananas" (LeetCode 875).
The same pattern: minimum capacity to ship packages in D days, minimum days to make bouquets, smallest divisor under a threshold, minimum time to paint boards.

Clarifying questions to ask

Is h at least the number of piles? (Otherwise it's impossible.)
Does leftover time in an hour carry over to another pile? (No.)
How large can the piles and h be? (Sums need 64-bit.)

Follow-ups to expect

"Why is binary search valid here?"
"What's the search range?"
"How do you compute hours for a speed?"
"Complexity?"
"Can you tighten the lower bound?"

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