How do you solve Minimum Number of Days to Make m Bouquets on LeetCode?

Minimum Number of Days to Make m Bouquets (LeetCode #1482) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log d) time complexity and O(1) space complexity. Key insight: If we can make m bouquets on day x, we can make them on any day after x.

What is the time complexity of Minimum Number of Days to Make m Bouquets?

The optimal solution for Minimum Number of Days to Make m Bouquets runs in O(n log d) time and uses O(1) space. The complexity is O(n log d) where n is the number of flowers and d is the maximum bloom day. We check each day in a binary search manner.

What is the brute force solution for Minimum Number of Days to Make m Bouquets?

The brute force approach for Minimum Number of Days to Make m Bouquets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each day to see if we can create the required number of bouquets. We simulate the blooming process day by day, counting how many bouquets can be formed. The optimal Optimal Solution approach improves this to O(n log d).

What algorithm or data structure is used to solve Minimum Number of Days to Make m Bouquets?

Minimum Number of Days to Make m Bouquets uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Greedy Algorithms.

What are the interview tips for Minimum Number of Days to Make m Bouquets?

Always clarify the problem constraints and edge cases before diving into coding. Think about how binary search can reduce the number of checks needed. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Number of Days to Make m Bouquets?

Not checking if the total number of flowers is less than m * k before proceeding. Failing to properly implement the bouquet counting logic.

#1482

Minimum Number of Days to Make m Bouquets

Medium

ArrayBinary SearchBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log d)
Space	O(1)	O(1)

💡

Intuition

Time O(n log d)Space O(1)

Using binary search allows us to efficiently find the minimum day by checking if we can create the required bouquets on that day. This reduces the number of days we need to check.

⚙️

Algorithm

4 steps

1Step 1: Check if it's possible to make m bouquets with k flowers each; if not, return -1.
2Step 2: Set up binary search between the minimum and maximum bloom days.
3Step 3: For each mid day in the binary search, check if we can make m bouquets.
4Step 4: Adjust the search range based on whether we can make the bouquets.

solution.py32 lines

1# Full working Python code
2
3def canMakeBouquets(bloomDay, m, k, days):
4    count = 0
5    bouquets = 0
6    for bloom in bloomDay:
7        if bloom <= days:
8            count += 1
9            if count == k:
10                bouquets += 1
11                count = 0
12        else:
13            count = 0
14    return bouquets >= m
15
16
17def minDays(bloomDay, m, k):
18    if len(bloomDay) < m * k:
19        return -1
20    left, right = 1, max(bloomDay)
21    while left < right:
22        mid = (left + right) // 2
23        if canMakeBouquets(bloomDay, m, k, mid):
24            right = mid
25        else:
26            left = mid + 1
27    return left
28
29bloomDay = [1,10,3,10,2]
30m = 3
31k = 1
32print(minDays(bloomDay, m, k))

ℹ

Complexity note: The complexity is O(n log d) where n is the number of flowers and d is the maximum bloom day. We check each day in a binary search manner.

1If we can make m bouquets on day x, we can make them on any day after x.
2We need to ensure that the total number of flowers is sufficient to meet the bouquet requirements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.