Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Koko Eating Bananas
Optimal Binary Search Approach



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 PilesHours (h)Minimum Eating Speed (k)Description
[3, 6, 7, 11]84Koko can eat all piles in 8 hours if she eats 4 bananas/hour
[30, 11, 23, 4, 20]530She must eat very fast to finish in just 5 hours
[30, 11, 23, 4, 20]623With 6 hours, speed of 23 is sufficient
[1, 1, 1, 1]41Minimum speed needed is 1 banana/hour
[1000000000]2500000000Single large pile, she must eat half in each hour
[]50No piles to eat, so minimum speed is 0
[10]110Only 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

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

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
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

CaseTime ComplexityExplanation
Best CaseO(n * log m)Where n = number of piles, m = max(pile). Binary search over m with O(n) check each time.
Average CaseO(n * log m)Performs log(m) iterations of binary search with O(n) check per iteration.
Average CaseO(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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