The given code attempts to solve the **"Koko Eating Bananas"**...

May 18, 2025 at 06:35 AM

public class Solution { public int minEatingSpeed(int[] piles, int h) { int l = 1; int r = Arrays.stream(piles).max().getAsInt(); int res = r; while (l <= r) { int k = (l + r) / 2; long totalTime = 0; for (int p : piles) { totalTime += Math.ceil((double) p / k); } if (totalTime <= h) { res = k; r = k - 1; } else { l = k + 1; } } return res; } }

The given code attempts to solve the "Koko Eating Bananas" problem, a popular algorithmic problem. Here's a detailed explanation of what it does:

Description of the Code

The method minEatingSpeed(int[] piles, int h) finds the minimum eating speed ( k ) (bananas per hour) such that Koko can eat all the bananas in the piles array within ( h ) hours.


Step-by-Step Explanation

  1. Binary Search Initialization:

    • l (left) is initialized as 1. This is the minimum possible speed ( k ), since Koko can eat at least 1 banana per hour.
    • r (right) is initialized as the maximum value in piles. This is the upper bound of the eating speed ( k ), as Koko could eat all bananas in the largest pile in one hour.
    • res stores the result (minimum speed ( k )) and is initially set to r.
  2. Binary Search for the Minimum Eating Speed ( k ):

    • While ( l \leq r ), the program uses binary search to calculate the mid-point speed ( k ): ( k = (l + r) / 2 ).
  3. Calculate Total Time to Eat Bananas at Speed ( k ):

    • For each pile in piles:
      • Compute the time required to finish the pile with the current speed ( k ). This is Math.ceil((double) p / k).
      • Sum up the times for all piles into totalTime.
  4. Update Binary Search Range Based on Total Time:

    • If totalTime <= h, it means Koko can eat all bananas within ( h ) hours at the current speed or less:
      • Update res = k (store the current minimum achievable speed).
      • Narrow the search to smaller speeds by setting r = k - 1.
    • Otherwise, if totalTime > h, it means ( k ) is too slow:
      • Increase the lower bound by setting l = k + 1.
  5. Return the Result:

    • The loop ends when the minimum speed is found, and the method returns res, which is the smallest ( k ) such that all bananas are eaten within ( h ) hours.

Key Concepts in the Code

  • Binary Search: Used to optimize the search for the minimum ( k ) efficiently in ( O(N \cdot \log(M)) ), where ( N ) is the number of piles and ( M ) is the size of the largest pile.
  • Mathematical Ceiling: The use of Math.ceil ensures that Koko finishes a pile even if the bananas do not divide evenly by ( k ).

Example Execution

Suppose piles = [3, 6, 7, 11] and h = 8:

  1. Initialize:

    • l = 1, r = 11 (max of piles), res = 11.
  2. Binary Search Steps:

    • ( k = (1 + 11) / 2 = 6 ): Compute totalTime at ( k = 6 ):

      • ceil(3/6) + ceil(6/6) + ceil(7/6) + ceil(11/6) = 1 + 1 + 2 + 2 = 6.
      • Since totalTime <= h, update res = 6, and set r = 5.
    • ( k = (1 + 5) / 2 = 3 ): Compute totalTime at ( k = 3 ):

      • ceil(3/3) + ceil(6/3) + ceil(7/3) + ceil(11/3) = 1 + 2 + 3 + 4 = 10.
      • Since totalTime > h, set l = 4.
    • ( k = (4 + 5) / 2 = 4 ): Compute totalTime at ( k = 4 ):

      • ceil(3/4) + ceil(6/4) + ceil(7/4) + ceil(11/4) = 1 + 2 + 2 + 3 = 8.
      • Since totalTime <= h, update res = 4, and set r = 3.
  3. Loop Ends:

    • Return res = 4.

Output

The minimum eating speed ( k ) is 4 bananas per hour.

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