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This approach uses binary search to efficiently determine the minimum number of days needed to make the required number of bouquets. By evaluating the feasibility of creating bouquets for a specific day, the search narrows down the smallest number of days possible.
The idea is to perform binary search between the minimum day and the maximum day in the bloomDay array. For each middle day value in the binary search, check if it's possible to make the required number of bouquets within those days. This check involves iterating through the bloomDay array and simulating the bouquet-making process.
If it's possible to make the bouquets on the current day or fewer, adjust the search range towards smaller values; otherwise, increase the range.
Time Complexity: O(n log(max(bloomDay))) where n is the length of the bloomDay array. This is due to binary searching over the day range and checking each day.
Space Complexity: O(1) since no additional data structures proportional to input size are used.
1def minDays(bloomDay, m, k):
2 def canMakeBouquets(days):
3 bouquets = 0
4 flowers = 0
5 for bloom in bloomDay:
6 if bloom <= days:
7 flowers += 1
8 if flowers == k:
9 bouquets += 1
10 flowers = 0
11 else:
12 flowers = 0
13 return bouquets >= m
14 if len(bloomDay) < m * k:
15 return -1
16 low, high = min(bloomDay), max(bloomDay)
17 while low < high:
18 mid = (low + high) // 2
19 if canMakeBouquets(mid):
20 high = mid
21 else:
22 low = mid + 1
23 return lowThe Python solution uses a binary search function that tests possible day numbers. The helper function 'canMakeBouquets' determines if bouquets can be made within a certain number of days by iterating through the bloomDay array and counting eligible flowers and bouquets.
A more straightforward but less optimal approach is to simulate each day sequentially and check whether that day allows the creation of the necessary bouquets. This simulation checks for the day on each flower to see if the flower is usable and forms bouquets if possible. This approach refines checking for the earliest feasible day without having an initial constraint of possible day ranges.
Time Complexity: O(n log(max(bloomDay))) as it checks each day incrementally
Space Complexity: O(1)
1#include <stdio.h>
2#include
This C solution iterates through each possible day, incrementing the number of bouquets built with adjacent flowers. It returns the minimum possible day when m bouquets can be successfully made, or -1 if impossible.