LeetCode #875 — MEDIUM

Koko Eating Bananas

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Koko loves to eat bananas. There are n piles of bananas, the i^th 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.

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

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[3,6,7,11]
8

Example 2

[30,11,23,4,20]
5

Example 3

[30,11,23,4,20]
6

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #875: Koko Eating Bananas
class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int l = 1, r = (int) 1e9;
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            for (int x : piles) {
                s += (x + mid - 1) / mid;
            }
            if (s <= h) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #875: Koko Eating Bananas
func minEatingSpeed(piles []int, h int) int {
	return 1 + sort.Search(slices.Max(piles), func(k int) bool {
		k++
		s := 0
		for _, x := range piles {
			s += (x + k - 1) / k
		}
		return s <= h
	})
}

# Accepted solution for LeetCode #875: Koko Eating Bananas
class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        def check(k: int) -> bool:
            return sum((x + k - 1) // k for x in piles) <= h

        return 1 + bisect_left(range(1, max(piles) + 1), True, key=check)

// Accepted solution for LeetCode #875: Koko Eating Bananas
impl Solution {
    pub fn min_eating_speed(piles: Vec<i32>, h: i32) -> i32 {
        let mut l = 1;
        let mut r = *piles.iter().max().unwrap_or(&0);
        while l < r {
            let mid = (l + r) >> 1;
            let mut s = 0;
            for x in piles.iter() {
                s += (x + mid - 1) / mid;
            }
            if s <= h {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}

// Accepted solution for LeetCode #875: Koko Eating Bananas
function minEatingSpeed(piles: number[], h: number): number {
    let [l, r] = [1, Math.max(...piles)];
    while (l < r) {
        const mid = (l + r) >> 1;
        const s = piles.map(x => Math.ceil(x / mid)).reduce((a, b) => a + b);
        if (s <= h) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.