
Problem Statement
Koko loves to eat bananas and has a unique way of doing so. She has access to n
piles of bananas where each pile contains a specific number of bananas denoted by piles[i]
. Koko has a limited time, as the guards who watch over the bananas will return in h
hours. She can set her own pace by choosing a rate k
, which represents how many bananas she can eat per hour. When Koko eats, she focuses on one pile per hour, consuming k
bananas from it. If a pile contains fewer than k
bananas, she will finish that pile and wait until the next hour to continue. Koko's challenge is to determine the minimal eating speed k
that allows her to consume all bananas across all piles within the time before the guards return.
Examples
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
Approach and Intuition
To solve the problem of finding the minimum eating speed k
for Koko, we need to grasp the relationship between k
and the total time T
taken to consume all the bananas:
Initial Observations:
- If
k
is very large (i.e., at least equal to the largest pile), Koko can finish the bananas in each pile in roughly one hour per pile or even less. - If
k
is very small, the number of hours required to finish all bananas increases significantly.
- If
Understanding the relationship:
- For a given
k
, the total hoursT
needed is the sum of the hours required for each pile, which is calculated as the ceiling of the division of pile size byk
(i.e., each pilepiles[i]
requires(piles[i] + k - 1) // k
hours).
- For a given
Binary Search for Efficiency:
- Given the constraints and the nature of the problem (searching for a minimum
k
that allows completing all piles withinh
hours), a binary search approach is ideal. - Set initial boundaries for
k
: the lower bound can be 1 (minimum possible speed), and the upper bound can be set to the maximum number inpiles
(maximum speed required if Koko decides to eat the largest pile in one hour). - We use a conditional loop to narrow down the value of
k
. In each iteration, choose the middle value ofk
, calculate the resultant hoursT
for thisk
, and adjust the bounds:- If
T
is less than or equal toh
, it meansk
might be optimized further; thus, reduce the upper bound. - If
T
is greater thanh
, increase the lower bound to explore higher values ofk
for feasibility.
- If
- Given the constraints and the nature of the problem (searching for a minimum
Finalizing the result:
- The iterative refinement via binary search will converge when the lower and upper bounds meet, resulting in the smallest
k
that allows Koko to finish all the bananas within the allottedh
hours.
- The iterative refinement via binary search will converge when the lower and upper bounds meet, resulting in the smallest
This approach efficiently handles the large possible range of values due to the constraints and directly addresses the question of the minimal satisfactory k
.
Solutions
- C++
- Java
- JavaScript
- Python
class Solution {
public:
int findMinSpeed(vector<int>& piles, int H) {
int low = 1, high = *max_element(piles.begin(), piles.end());
while (low < high) {
int mid = (low + high) / 2;
int totalHours = 0;
for (int bananas : piles) {
totalHours += bananas / mid + (bananas % mid != 0);
}
if (totalHours <= H) {
high = mid;
} else {
low = mid + 1;
}
}
return low;
}
};
In the given C++ code, the solution focuses on solving the problem where Koko can eat bananas at a variable rate but must finish all the bananas within a set number of hours, H. The essence of the solution leverages a binary search approach to efficiently find the minimum integer rate at which Koko should eat the bananas to achieve this goal.
Here's a breakdown of how the solution works:
Start by setting the initial search range for the eating rate (
low
tohigh
). Thelow
is set to 1 (minimum possible eating rate) andhigh
is set to the maximum number of bananas in any pile using themax_element
function. This ensures the rate is logically bounded by the maximum pile size, as Koko cannot eat faster than the size of the largest pile per hour.Execute a binary search:
- Calculate a mid-point rate (
mid
) to test. - For each pile of bananas, compute how many hours it would take for Koko to eat that pile at the
mid
rate, rounding up to ensure each pile is fully eaten. - Sum these hours across all piles.
- Calculate a mid-point rate (
Determine if the computed total hours fit within the allowed hours (
H
):- If Koko can finish within
H
hours at themid
rate, search for a potentially lower rate by adjustinghigh
tomid
. - If more hours are needed, adjust
low
tomid + 1
to try a faster eating rate.
- If Koko can finish within
The search concludes when the
low
meetshigh
, at which pointlow
will represent the minimal eating rate that allows Koko to finish all the bananas withinH
hours.
By the conclusion of this code execution, you obtain the minimized eating rate as a result, enabling efficient completion of the task within the set constraints.
class Solution {
public int minimumEatingSpeed(int[] piles, int hours) {
int minSpeed = 1, maxSpeed = 1;
for (int pile : piles) {
maxSpeed = Math.max(maxSpeed, pile);
}
while (minSpeed < maxSpeed) {
int mid = (minSpeed + maxSpeed) / 2;
int totalHours = 0;
for (int pile : piles) {
totalHours += Math.ceil((double) pile / mid);
}
if (totalHours <= hours) {
maxSpeed = mid;
} else {
minSpeed = mid + 1;
}
}
return maxSpeed;
}
}
The Java code provided solves the problem of determining the minimum eating speed at which Koko can eat all the bananas within the given hours. The code uses a binary search approach. This method efficiently narrows down the speed range until it finds the minimum speed that allows Koko to finish the bananas in time.
- The
minimumEatingSpeed
function takes two parameters: an arraypiles
of banana counts and an integerhours
for the maximum allowed time. - Initialize
minSpeed
to 1 andmaxSpeed
to the maximum element inpiles
found using a loop, as Koko cannot eat at a speed slower than 1 or faster than the largest pile per hour. - Execute a binary search between
minSpeed
andmaxSpeed
:- Calculate the mid-point
mid
representing a potential eating speed. - Calculate total hours required to eat all piles at this speed. Round up the division result for each pile to consider partial hours as full hours.
- If
totalHours
is less than or equal tohours
, adjustmaxSpeed
tomid
because a lower or equal speed might also be a valid solution. - Otherwise, increase
minSpeed
tomid + 1
to look for a faster valid speed as the current one is too slow.
- Calculate the mid-point
- When the loop terminates,
maxSpeed
holds the minimum speed at which Koko can eat all the bananas within the given hours.
This approach optimally balances between too high and too low speeds and ensures that the solution is both correct and efficient, typically achieving O(n log m) complexity, where n is the number of piles, and m is the range of possible speeds.
var minimumEatingSpeed = function(bananas, maxHours) {
let low = 1, high = 1;
for (const banana of bananas) {
high = Math.max(high, banana);
}
while (low < high) {
let mid = Math.floor((low + high) / 2);
let totalHours = 0;
for (const banana of bananas) {
totalHours += Math.ceil(banana / mid);
}
if (totalHours <= maxHours) {
high = mid;
} else {
low = mid + 1;
}
}
return low;
}
The code provided defines a function in JavaScript intended to solve the problem of finding the minimum speed at which Koko must eat bananas in order to consume all bananas within a given maximum number of hours. This problem context involves figuring out an optimal eating rate given constraints on time and quantity.
- In the function
minimumEatingSpeed
, two parameters are expected:bananas
, an array where each element represents the number of bananas in a pile, andmaxHours
, which depicts the maximum hours Koko has to eat all the bananas. - To find the solution, the function employs a binary search technique. It initializes two variables,
low
andhigh
.low
starts at 1 (since the minimum possible eating speed is at least one banana per hour), andhigh
is set initially to the maximum number of bananas in any pile, ensuring the search boundary encompasses all possible eating speeds from slowest to fastest required.
The main logic:
- A loop computes the values for
high
to ensure it captures the largest pile count from thebananas
array usingMath.max
. - Through the while loop, the binary search continues until
low
is less thanhigh
. - Within this loop,
mid
represents the middle value betweenlow
andhigh
, suggesting a potential answer for the minimum eating speed. - Another loop calculates
totalHours
by iterating through every pile; it calculates the hours it would take for Koko to eat each pile atmid
speed usingMath.ceil
to account for partial hours. - Decision-making inside the while loop adjusts the bounds (
low
andhigh
) based on whether the calculatedtotalHours
with the current mid value is within the allowablemaxHours
. - If Koko can finish the bananas in fewer or equal hours than
maxHours
, reduce the upper boundary (high = mid
), otherwise increase the lower boundary (low = mid + 1
).
At the end of the binary search, low
holds the minimum speed at which Koko must eat the bananas to meet the time constraint, and this value is returned by the function. This approach efficiently handles the constraints and finds an optimal rate using a time complexity typically logarithmic relative to the initial high
value setup, providing a quick and scalable solution.
class Solution:
def findMinSpeed(self, piles: List[int], hours: int) -> int:
# Define search range
low = 1
high = max(piles)
# Using binary search to find the optimal speed
while low < high:
mid = (low + high) // 2
total_hours = 0
# Calculate total hours taken to finish all piles with current speed
for pile in piles:
total_hours += math.ceil(pile / mid)
# Narrow down the search range based on the number of hours
if total_hours <= hours:
high = mid
else:
low = mid + 1
# low or high is the answer
return low
In the provided Python solution, the goal is to determine the minimum eating speed at which Koko can eat all the banana piles within a given number of hours. The function findMinSpeed
employs a binary search mechanism to efficiently find this speed.
Here's how the solution works:
Begin by defining the range for Koko's possible eating speeds. This ranges from
1
to the maximum number of bananas in any pile since Koko must eat at least one banana per hour and cannot eat faster than the size of the largest pile per hour.The search for the optimal speed is a classic binary search where:
- Calculate the middle of the current range (
mid
) as the potential eating speed. - Compute the total hours required to eat all piles at this speed. For each pile, use the ceiling of the division of pile size by current mid (i.e.,
math.ceil(pile / mid)
), which ensures that partial bananas still require a full hour to complete.
- Calculate the middle of the current range (
Adjust the search range based on whether the computed total hours exceed the allowed hours:
- If the total hours needed are less than or equal to the allowed hours, then reduce the upper boundary of the search range (
high
) tomid
. - If the total hours exceed the allowed hours, increase the lower boundary of the search range (
low
) tomid + 1
.
- If the total hours needed are less than or equal to the allowed hours, then reduce the upper boundary of the search range (
When
low
meets or exceedshigh
, the search concludes, andlow
is the minimum speed at which Koko can finish all the bananas within the allowed hours.
This search leverages the efficiency of binary search to find the solution in logarithmic time relative to the range of speeds, making it very efficient even for large inputs. The use of math.ceil
ensures that the calculation accounts for partially eaten piles correctly for every hour.
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