### bablu_45's blog

By bablu_45, history, 15 months ago, Given an array of roses. roses[i] means rose i will bloom on day roses[i]. Also given an int k, which is the minimum number of adjacent bloom roses required for a bouquet, and an int n, which is the number of bouquets we need. Return the earliest day that we can get n bouquets of roses.

Example: Input: roses = [1, 2, 4, 9, 3, 4, 1], k = 2, n = 2

Output: 4

Explanation:

day 1: [b, n, n, n, n, n, b]. The first and the last rose bloom.

day 2: [b, b, n, n, n, n, b]. The second rose blooms. Here the first two bloom roses make a bouquet.

day 3: [b, b, n, n, b, n, b]

day 4: [b, b, b, n, b, b, b]

Here the last three bloom roses make a bouquet, meeting the required n = 2 bouquets of bloom roses. So return day 4.

I am looking for a efficient D.P solution (if possible). Comments (5)
 » Do a binary search on number of days. the complexity would be O(log(num_days)) * O(num_roses)
•  » » 15 months ago, # ^ | ← Rev. 3 →   I am sorry that I forgot to mention that I was looking for an efficient D.P solution (if possible) even if it's worse than binary search.Thanks anyway.
 » $O(n)$ solution:Wlog suppose the array has length which is a multiple of $k$. Decompose the array into blocks of length $k$, and in each block, compute prefix maximums and suffix maximums. Now, for any window of length $k$, we can query the maximum in that window in $O(1)$ time by combining a suffix and prefix of two adjacent blocks. So, iterate over all windows of length $k$, and find the window with smallest maximum. That's the answer.This is a common technique known as sliding window RMQ or fixed-range RMQ.
•  » » That would be correct if you wanted to make only one bouquet, but I think you need to make n bouqets...Correct me if I am wrong.
•  » » » Oh, you're completely right, I didn't read the problem correctly.Of course the binary search solution still works in $O(m\log(\max(\text{roses})))$. (Let $m$ be the length of the original array). We can use compression to get $O(m\log m)$ as well. I don't think it can be solved any faster than that for $n\neq 1$.