The given code attempts to solve the **"Koko Eating Bananas"**...
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
-
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 inpiles
. 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 tor
.
-
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 ).
-
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
.
- Compute the time required to finish the pile with the current speed ( k ). This is
- For each pile in
-
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
.
- Update
- Otherwise, if
totalTime > h
, it means ( k ) is too slow:- Increase the lower bound by setting
l = k + 1
.
- Increase the lower bound by setting
- If
-
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.
- The loop ends when the minimum speed is found, and the method returns
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
:
-
Initialize:
l = 1
,r = 11
(max ofpiles
),res = 11
.
-
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
, updateres = 6
, and setr = 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
, setl = 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
, updateres = 4
, and setr = 3
.
-
-
Loop Ends:
- Return
res = 4
.
- Return
Output
The minimum eating speed ( k ) is 4
bananas per hour.