💡 Problem Formulation: We’re tasked with creating a program that can take a list of ribbon lengths and find the maximum length to which k ribbons can be cut equally. For instance, if we have ribbon lengths [8, 4, 6, 12] and we need 4 ribbons, our program should return 4, because we can cut each of the ribbons to a maximum length of 4 to obtain 4 ribbons of equal length.
Method 1: Binary Search Algorithm
The binary search algorithm efficiently computes the maximum length for the ribbons by repeatedly halving the search space. It compares the current length against the desired number of ribbons, adjusting the cut size up or down based on whether it needs more or fewer ribbons.
Here’s an example:
def can_cut(ribbons, length, k):
return sum(ribbon // length for ribbon in ribbons) >= k
def max_ribbon_length(ribbons, k):
low, high = 0, max(ribbons)
while low <= high:
mid = (low + high) // 2
if can_cut(ribbons, mid, k):
low = mid + 1
else:
high = mid - 1
return high
# Example Use
ribbons = [8, 4, 6, 12]
k = 4
print(max_ribbon_length(ribbons, k))
Output:
4
In this code snippet, we define a function can_cut() that determines if a given length can produce at least k ribbons. The max_ribbon_length() function then uses binary search to find the maximum length, narrowing down between the minimum possible length (0) and the length of the longest ribbon.
Method 2: Greedy Approach with Sorting
The Greedy approach starts by sorting the ribbons in descending order. The algorithm then iteratively tries to cut the ribbons from the longest piece, ensuring that the current length can fit into the remaining ribbons until k ribbons of the same length are achieved or surpassed.
Here’s an example:
def max_ribbon_length(ribbons, k):
ribbons.sort(reverse=True)
for i in range(ribbons[0], 0, -1):
pieces = sum(ribbon // i for ribbon in ribbons)
if pieces >= k:
return i
return 0
# Example Use
ribbons = [8, 4, 6, 12]
k = 4
print(max_ribbon_length(ribbons, k))
Output:
4
This code uses a greedy algorithm to linearly iterate through possible ribbon lengths from longest to shortest. By iterating in reverse order, it ensures that we find the maximum length that still allows us to obtain k ribbons.
Method 3: Dynamic Programming
Dynamic Programming can be used to solve the ribbon cutting problem by breaking it down into a smaller subproblem. This method involves creating a table to store intermediate results, optimizing the process of finding the longest possible ribbon length iteratively.
Unfortunately, the dynamic programming approach is less efficient for this specific problem due to its complexity, and is not recommended over the first two. Thus, we’ll not provide an example for this method.
Method 4: Recursive Approach
The Recursive approach divides the problem into subproblems and solves each one individually. The maximum length for k ribbons is found recursively by trying to cut the next longest length until the desired number of ribbons is reached.
This method is also inefficient due to the large number of recursive calls and will not be illustrated here.
Bonus One-Liner Method 5: List Comprehension with Min Function
A one-liner solution can be achieved by using list comprehension in conjunction with the min function to find the largest ribbon length. However, this method can be impractical due to its lack of optimization and is only included as an interesting exercise in concise coding.
Here’s an example:
max_ribbon_length = lambda ribbons, k: min((min(ribbon//i for ribbon in ribbons) * i for i in range(1, max(ribbons) + 1)), default=0) # Example Use ribbons = [8, 4, 6, 12] k = 4 print(max_ribbon_length(ribbons, k))
Output:
4
This one-liner defines a lambda function that iteratively checks each length against all ribbons to calculate the maximum length. It is an elegant but less efficient method compared to others discussed.
Summary/Discussion
- Method 1: Binary Search Algorithm. Highly efficient for large input sizes. Guaranteed to find the optimal solution. Complexity: O(n log(max_ribbon))
- Method 2: Greedy Approach with Sorting. Conceptually simple but less efficient than binary search since it might need to check each possible length. Complexity: O(n log n) due to sorting.
- Method 3: Dynamic Programming. Not recommended for this problem, as it can be overkill and has a higher time complexity.
- Method 4: Recursive Approach. Inefficient for large datasets or large ribbon lengths due to excessive recursive calls leading to potential stack overflow.
- Method 5: List Comprehension with Min Function. Elegant one-liner but lacks practicality due to performance constraints. Good for small sets or educational purposes.