Koko Eating Bananas
Optimal Binary Search Approach

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Problem Statement

Koko the monkey loves bananas. She has a list of banana piles, and she wants to eat all of them within h hours. Each hour, Koko chooses any pile and eats up to k bananas from it. If the pile has fewer than k bananas, she eats the entire pile and moves on.

Your task is to help Koko determine the minimum integer eating speed k such that she can finish all the bananas in at most h hours.

If the list of banana piles is empty, return 0 as there are no bananas to eat.

Examples❯

Banana Piles	Hours (h)	Minimum Eating Speed (k)	Description
[3, 6, 7, 11]	8	4	Koko can eat all piles in 8 hours if she eats 4 bananas/hour
[30, 11, 23, 4, 20]	5	30	She must eat very fast to finish in just 5 hours
[30, 11, 23, 4, 20]	6	23	With 6 hours, speed of 23 is sufficient
[1, 1, 1, 1]	4	1	Minimum speed needed is 1 banana/hour
[1000000000]	2	500000000	Single large pile, she must eat half in each hour
[]	5	0	No piles to eat, so minimum speed is 0
[10]	1	10	Only one pile, must eat all in one hour

Solution❯

This problem can feel tricky at first, but it becomes easier once you understand what you're really being asked: "What's the slowest speed Koko can eat at and still finish all the bananas in h hours or less?"

The idea is to check different eating speeds and simulate how many hours Koko would take at each one. The smaller the speed, the longer it takes. The larger the speed, the faster she finishes. So we’re looking for the minimum speed that still allows her to finish on time.

What Happens at a Certain Speed?

Let’s say we try a speed of k bananas per hour. For every pile, we calculate how many hours Koko would take to eat that pile: it's ceil(pile / k) hours. We add up these hours for all piles. If the total time is less than or equal to h, then this speed k works — but maybe there’s a slower speed that also works, so we try smaller values. If the total time is more than h, then she’s eating too slow and we need to try higher speeds.

How Do We Find the Best Speed Efficiently?

Instead of testing every possible speed from 1 to the maximum pile size (which could be huge), we use binary search. This helps us quickly zero in on the minimum valid speed. Here's the intuition:

low starts at 1 — the slowest possible eating speed.
high is the largest pile — the fastest she'd ever need to eat (if there's just one hour).
We pick a mid speed between low and high and check if it’s fast enough.
If it is, we try to find a slower valid speed (search left); if not, we try faster (search right).

Different Cases to Think About

Exact Fit: When h is equal to the total number of piles, Koko must eat at least one pile per hour. So her speed must be high enough to finish any one pile in one hour.
Relaxed Time: If h is very large, she can afford to eat slowly.
Only One Pile: The answer is simply ceil(pile / h).
Empty Piles List: If there are no piles, Koko doesn’t need to eat. So the answer is 0.

By adjusting the speed intelligently using binary search, we quickly find the minimum speed at which she can still meet her goal.

Visualization

Algorithm Steps

Initialize low = 1, high = max(piles), and answer = max(piles).
While low ≤ high:
→ Compute mid = (low + high) / 2 as current eating speed.
→ For each pile, calculate hours = ceil(pile / mid).
→ Sum total hours for all piles.
→ If total hours ≤ h: update answer = mid, and search left by setting high = mid - 1.
→ Else, search right with low = mid + 1.
Return answer as the minimum valid eating speed.

Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

import math

def minEatingSpeed(piles, h):
    low, high = 1, max(piles)  # Possible eating speeds range from 1 to max pile
    answer = high

    while low <= high:
        mid = (low + high) // 2
        hours = sum(math.ceil(p / mid) for p in piles)  # Total hours at speed 'mid'

        if hours <= h:
            answer = mid       # Try a smaller speed
            high = mid - 1
        else:
            low = mid + 1      # Need more speed to eat faster

    return answer

# Example
print(minEatingSpeed([3, 6, 7, 11], 8))  # Output: 4

Time Complexity

Case	Time Complexity	Explanation
Best Case	O(n * log m)	Where n = number of piles, m = max(pile). Binary search over m with O(n) check each time.
Average Case	O(n * log m)	Performs log(m) iterations of binary search with O(n) check per iteration.
Worst Case	O(n * log m)	In worst case, binary search runs full range from 1 to max(pile), checking all piles each time.

Space Complexity

O(1)

Explanation: Only a constant number of variables are used for computation. No additional memory is required.

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Sorting6
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Arrays32
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Arrays - Binary Search28
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Strings18
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Matrix3
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Matrix - Binary Search5
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Binary Trees36
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Binary Search Trees14
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Graphs46
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Tries1
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Koko Eating Bananas
Optimal Binary Search Approach

Problem Statement

Examples❯

Solution❯

What Happens at a Certain Speed?

How Do We Find the Best Speed Efficiently?

Different Cases to Think About

Visualization

Algorithm Steps

Code

Time Complexity

Space Complexity

Support ProgramGuru.org❯

Koko Eating BananasOptimal Binary Search Approach

Problem Statement

Examples❯

Solution❯

What Happens at a Certain Speed?

How Do We Find the Best Speed Efficiently?

Different Cases to Think About

Visualization

Algorithm Steps

Code

Time Complexity

Space Complexity

Welcome to ProgramGuru

Support ProgramGuru.org❯

Koko Eating Bananas
Optimal Binary Search Approach