MEDIUM NC#30 Binary Search
875. Koko Eating Bananas
📖 Problem
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. Return the minimum integer k such that she can eat all the bananas within h hours.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Midpoint overflow-safe math
- •Loop invariants
- •Monotonic condition design
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Binary Search reasoning pattern
- •Apply Backtracking reasoning pattern
💡 Approach
- → Use binary search to find minimum eating speed k
- → Lower bound: 1, Upper bound: max(piles)
- → For each k, calculate hours needed: sum(ceil(pile/k))
- → If hours <= h, try smaller k; else try larger k
- → Time: O(n * log(max(piles))), Space: O(1)
🧭 Prerequisites
🛠️ Hints & Pitfalls
Hints
- •Use binary search to find minimum eating speed k
- •Lower bound: 1, Upper bound: max(piles)
- •For each k, calculate hours needed: sum(ceil(pile/k))
Common Pitfalls
- •If hours <= h, try smaller k; else try larger k
- •Time: O(n * log(max(piles))), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 2
Test Case 1
Not run Input:
minEatingSpeed([3,6,7,11], 8); Expected:
4 Test Case 2
Not run Input:
minEatingSpeed([30,11,23,4,20], 5); Expected:
30 Test Case 3
Not run Input:
minEatingSpeed([30,11,23,4,20], 6); Expected:
23 📝 Code Editor
📤 Output