Watch 10 video solutions for Minimum Number of Days to Make m Bouquets, a medium level problem involving Array, Binary Search. This walkthrough by codestorywithMIK has 16,816 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer array bloomDay, an integer m and an integer k.
You want to make m bouquets. To make a bouquet, you need to use k adjacent flowers from the garden.
The garden consists of n flowers, the ith flower will bloom in the bloomDay[i] and then can be used in exactly one bouquet.
Return the minimum number of days you need to wait to be able to make m bouquets from the garden. If it is impossible to make m bouquets return -1.
Example 1:
Input: bloomDay = [1,10,3,10,2], m = 3, k = 1 Output: 3 Explanation: Let us see what happened in the first three days. x means flower bloomed and _ means flower did not bloom in the garden. We need 3 bouquets each should contain 1 flower. After day 1: [x, _, _, _, _] // we can only make one bouquet. After day 2: [x, _, _, _, x] // we can only make two bouquets. After day 3: [x, _, x, _, x] // we can make 3 bouquets. The answer is 3.
Example 2:
Input: bloomDay = [1,10,3,10,2], m = 3, k = 2 Output: -1 Explanation: We need 3 bouquets each has 2 flowers, that means we need 6 flowers. We only have 5 flowers so it is impossible to get the needed bouquets and we return -1.
Example 3:
Input: bloomDay = [7,7,7,7,12,7,7], m = 2, k = 3 Output: 12 Explanation: We need 2 bouquets each should have 3 flowers. Here is the garden after the 7 and 12 days: After day 7: [x, x, x, x, _, x, x] We can make one bouquet of the first three flowers that bloomed. We cannot make another bouquet from the last three flowers that bloomed because they are not adjacent. After day 12: [x, x, x, x, x, x, x] It is obvious that we can make two bouquets in different ways.
Constraints:
bloomDay.length == n1 <= n <= 1051 <= bloomDay[i] <= 1091 <= m <= 1061 <= k <= nProblem Overview: You are given an array bloomDay where each value represents the day a flower blooms. The goal is to make m bouquets, each requiring k adjacent flowers. The task is to return the minimum number of days required so that at least m bouquets can be formed.
Approach 1: Greedy Sequential Day Simulation (O(n * D) time, O(1) space)
This approach simulates the garden day by day. Start from day 1 and keep increasing the day until the required bouquets can be formed. For each day, iterate through the array and count consecutive flowers whose bloom day is less than or equal to the current day. Every time you reach k consecutive flowers, increment the bouquet count and reset the counter. If the bouquet count reaches m, that day is the answer. While easy to implement, the algorithm may check many days up to the maximum bloom day, making it inefficient for large ranges.
Approach 2: Binary Search on Days (O(n log D) time, O(1) space)
The key insight: if it is possible to make m bouquets on day d, then it will also be possible on any later day. This monotonic property makes the problem ideal for binary search. Search the answer in the range between the minimum and maximum bloom days. For each candidate day, run a greedy check by scanning the array and counting consecutive bloomed flowers. Whenever k adjacent flowers are available, form a bouquet and reset the counter. If you can form at least m bouquets, move the search left to minimize the day; otherwise move right.
This approach avoids checking every day and instead narrows the search range logarithmically. The bouquet counting step is essentially a greedy scan because you always form a bouquet as soon as k adjacent flowers are available.
Recommended for interviews: Binary Search on Days is the expected solution. The brute-force simulation demonstrates the problem mechanics, but the optimized binary search shows that you recognized the monotonic feasibility condition and reduced the search space efficiently.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Greedy Sequential Day Simulation | O(n * D) | O(1) | Useful for understanding the problem mechanics or when the day range is very small |
| Binary Search on Days | O(n log D) | O(1) | Optimal approach for large inputs where bloom days span a large range |