Understanding the Problem
We are given an array of integers where each element represents the day a flower will bloom. Our task is to make m
bouquets, each consisting of exactly k
adjacent bloomed flowers. A flower can only be used on or after the day it blooms.
The main question is: What is the minimum number of days needed to make m
such bouquets?
Step-by-Step Solution with Example
Step 1: Think About a Single Day
Let’s say we choose a day, like day 7. On this day, only flowers that bloom on or before day 7 can be used. We scan the array and try to form bouquets using only these bloomed flowers. We can only form a bouquet if we find k
adjacent bloomed flowers.
Step 2: Try to Simulate It
Let’s take an example:
flowers = [1, 10, 3, 10, 2], m = 1, k = 2
This means we need to make 1 bouquet using 2 adjacent bloomed flowers. On day 3, the bloom status is:
- Day 1 → bloomed (1 ≤ 3)
- Day 10 → not bloomed (10 > 3)
- Day 3 → bloomed (3 ≤ 3)
- Day 10 → not bloomed
- Day 2 → bloomed (2 ≤ 3)
The bloomed flowers are at index 0, 2, and 4. None of them are adjacent, so we cannot form any bouquet on day 3. Try day 10 — now all flowers are bloomed, and we can form a bouquet with [3, 10] or [10, 2]. So day 10 works. We try to find the minimum such day.
Step 3: Use Binary Search to Optimize
Since trying all days linearly is slow, we use binary search on days from the minimum to the maximum bloom value.
At each mid-point day, we check if it's possible to make m
bouquets. If it is, we try a smaller day. If it’s not, we try a bigger day. This helps us zero in on the minimum valid day efficiently.
Step 4: Bouquet Formation Check Logic
While scanning, we count how many adjacent bloomed flowers we have. If a flower is bloomed on or before the current day, we increase the count. If not, we reset the count to 0. Every time the count reaches k
, we form a bouquet and reset the count.
Edge Cases
- Not Enough Flowers: If
m × k
is greater than the total number of flowers, it's impossible to form the required bouquets. We should return -1
immediately.
- All Flowers Bloom Late: Even if all flowers bloom on high days (like day 1000), binary search will still find the answer quickly.
- Empty Array: If the array is empty, we cannot form any bouquet. Return
-1
.
- Exactly Enough Flowers: If the array has exactly
m × k
flowers, we must use every flower — and they must be arranged in proper adjacent groups. This is a tight case to test our adjacency logic.
Finally
We solved this problem using a binary search approach, which significantly improves performance over a brute force solution. The key insights were:
- Understanding how to simulate bouquet formation on a specific day
- Applying binary search on the range of days to minimize the answer
- Carefully handling edge cases, especially when adjacency is required
The time complexity is O(n × log(max bloom day)), where n
is the number of flowers. This makes the solution both fast and scalable.
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