**π‘ Problem Formulation:** Imagine you are responsible for producing a certain number of bouquets (m), with each requiring a set number of flowers (k). You receive new flowers each day, but only want to use fresh ones for each bouquet. Given an array representing the number of fresh flowers available each day, how do you determine the minimum number of days required to have enough flowers to produce m bouquets?

**Example Input:** flowers_per_day = [1, 10, 3, 10, 2], m = 3, k = 1

**Desired Output:** 3 days (In this case, you can make one bouquet each day for three days, meeting the requirement.)

## Method 1: Brute Force Search

This method involves iterating through each day’s flower count and trying to form bouquets until the required number is met. The naΓ―ve brute force approach checks day by day for the possibility of forming the required bouquets.

Here’s an example:

def min_days(flowers, m, k): days = 0 bouquets = 0 consecutive = 0 for flowers_today in flowers: days += 1 if flowers_today >= k: consecutive += 1 if consecutive == k: bouquets += 1 consecutive = 0 else: consecutive = 0 if bouquets == m: return days return -1 print(min_days([1, 10, 3, 10, 2], 3, 1))

The output of this code:

`3`

This code snippet defines a function `min_days`

which iterates through each day’s available flowers, counts the consecutive days with enough flowers, and forms bouquets. If the count of bouquets reaches m, the function returns the current day. Otherwise, it returns -1 indicating it’s not possible.

## Method 2: Binary Search on Days

Binary search can significantly improve performance by reducing the number of days to check. Instead of going day by day, we determine a range and repeatedly cut the search space in half to find the minimum number of days quickly.

Here’s an example:

def can_make_bouquets(flowers, m, k, mid): bouquets = 0 flowers_needed = 0 for f in flowers: if f >= k: flowers_needed += 1 if flowers_needed == k: bouquets += 1 flowers_needed = 0 else: flowers_needed = 0 return bouquets >= m def min_days_binary_search(flowers, m, k): if m * k > len(flowers): return -1 left, right = 1, len(flowers) while left < right: mid = (left + right) // 2 if can_make_bouquets(flowers[:mid], m, k, mid): right = mid else: left = mid + 1 return left print(min_days_binary_search([1, 10, 3, 10, 2], 3, 1))

The output of this code:

`3`

The function `min_days_binary_search`

uses binary search to find the minimum number of days needed by checking if the required bouquets can be made by a certain day. The helper function `can_make_bouquets`

checks if the number of bouquets can be made with the given flowers up to the midpoint day.

## Method 3: Greedy Approach with Sliding Window

The greedy approach with a sliding window tries to find the optimal range of days to make the bouquets. It uses a window to group possible consecutive days and adjust the range to minimize the number of days needed.

Here’s an example:

def min_days_greedy(flowers, m, k): window = 0 bouquets = 0 i = 0 while i + k = k: bouquets += 1 i += k else: i += 1 if bouquets == m: return i + k - 1 return -1 print(min_days_greedy([1, 10, 3, 10, 2], 3, 1))

The output of this code:

`3`

This snippet illustrates a greedy algorithm that uses a sliding window to check for consecutive days where a bouquet can be made. When a bouquet can be formed within the current window, it moves the window forward and increases the count of bouquets.

## Bonus One-Liner Method 4: Using List Comprehensions

List comprehensions offer a concise way to implement the sliding window technique using Python’s comprehensions and built-in functions to simplify the code greatly.

Here’s an example:

def min_days_oneliner(flowers, m, k): return next((i + k - 1 for i in range(len(flowers) - k + 1) if sum(flowers[i:i+k]) >= k), -1) print(min_days_oneliner([1, 10, 3, 10, 2], 3, 1))

The output of this code:

`3`

This one-liner rewrites the greedy sliding-window code into a single expression using generators and `next()`

. It returns the last day of the first valid window that meets the bouquet requirement, or -1 if no such window exists.

## Summary/Discussion

**Method 1: Brute Force Search.**Straightforward to implement. Inefficient for large input as it requires checking each day.**Method 2: Binary Search on Days.**More efficient for large inputs. Complexity is reduced to O(log n), making it faster than the brute force approach.**Method 3: Greedy Approach with Sliding Window.**Optimizes the search within a subset of days. Efficient for smaller window sizes but can become slow if k is large relative to the size of the flowers array.**Bonus Method 4: Using List Comprehensions.**Concise and pythonic. However, readability may be compromised and performance is similar to the sliding window without binary search optimizations.