Minimum Number Of Days To Make M Bouquets - Leetcode Solution

The problem is to find the minimum number of days it will take to make m bouquets given an array of integer roses, representing the number of roses in each day's bouquet. Each bouquet must contain exactly k roses and can only be made using consecutive days.

The brute force approach would be to try every possible combination of consecutive days and see which combination yields the minimum number of days to make m bouquets, which would take O(n^k) time complexity. However, we can use a binary search approach to optimize this problem.

We know that the minimum number of days to make m bouquets will be between the minimum and maximum number of roses in the array. We can perform a binary search on this range and check if it is possible to make m bouquets using k roses each day within a certain number of days.

We can keep track of the number of bouquets we can make in the current range of days and use this to adjust our search range accordingly. If we can make m bouquets within the current range of days, we check if we can do it within fewer days. Otherwise, we adjust our range to include more days and try again until we find the minimum number of days to make m bouquets.

Here's the step-by-step solution:

1. Set the left and right bounds of the binary search to the minimum and maximum number of roses in the array, respectively.

2. While left <= right, do the following:

   a. Calculate the mid value as (left + right) / 2.

   b. Initialize a bouquet counter and a flower counter to 0.

   c. Iterate through the array and for each day, check if the number of roses is less than or equal to mid. If it is, increment the flower counter. If the flower counter reaches k, increment the bouquet counter and reset the flower counter to 0.

   d. If the bouquet counter is less than m, we need to increase the number of days, so we set left to mid + 1 and continue the search.

   e. Otherwise, the bouquet counter is greater than or equal to m, so we check if it is possible to make m bouquets within fewer days. To do this, we set right to mid - 1 and continue the search.

3. Once the binary search is complete, return the left value, which represents the minimum number of days to make m bouquets.

The time complexity of this approach is O(n log r), where r is the range of possible days (maximum number of roses - minimum number of roses). The space complexity is O(1) as we only need to store a few integer variables.

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