Minimum Number of Days to Make m Bouquets

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 i<sup>th</sup> 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.

def minDays(bloomDay, m, k):
    """Finds the minimum number of days to wait to make m bouquets.

    Args:
        bloomDay (list[int]): An integer array representing the bloom day of each flower.
        m (int): The number of bouquets to make.
        k (int): The number of adjacent flowers needed for each bouquet.

    Returns:
        int: The minimum number of days to wait, or -1 if impossible.
    """
    n = len(bloomDay)

    if m * k > n:
        return -1

    def possible(days):
        bouquets = 0
        flowers = 0
        for bloom in bloomDay:
            if bloom <= days:
                flowers += 1
                if flowers == k:
                    bouquets += 1
                    flowers = 0
            else:
                flowers = 0
        return bouquets >= m

    left, right = 1, max(bloomDay)
    ans = right

    while left <= right:
        mid = (left + right) // 2
        if possible(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans

# Example usage
bloomDay1 = [1, 10, 3, 10, 2]
m1 = 3
k1 = 1
print(f"Example 1: {minDays(bloomDay1, m1, k1)}")  # Output: 3

bloomDay2 = [1, 10, 3, 10, 2]
m2 = 3
k2 = 2
print(f"Example 2: {minDays(bloomDay2, m2, k2)}")  # Output: -1

bloomDay3 = [7, 7, 7, 7, 12, 7, 7]
m3 = 2
k3 = 3
print(f"Example 3: {minDays(bloomDay3, m3, k3)}")  # Output: 12

Naive Solution

A naive approach would involve iterating through each day from the minimum bloom day to the maximum bloom day and, for each day, checking if it's possible to create m bouquets of k adjacent flowers. This approach is inefficient due to its high time complexity.

Optimal Solution

The optimal solution uses binary search to find the minimum number of days. We define a possible function that checks if it's possible to make m bouquets given a certain number of days. The binary search narrows down the range of days until the minimum number of days is found.

Big(O) Run-Time Analysis

The possible function iterates through the bloomDay array once, which takes O(n) time, where n is the length of bloomDay. The binary search runs in O(log(max(bloomDay))) time, where max(bloomDay) is the maximum bloom day. Therefore, the overall time complexity is O(n * log(max(bloomDay))).

Big(O) Space Usage Analysis

The space complexity is O(1) because the algorithm uses a constant amount of extra space, regardless of the input size. The possible function uses only a few variables to keep track of the number of bouquets and flowers, and the binary search uses a few variables for the left, right, and mid pointers.

Edge Cases

1. Impossible to make bouquets: If m * k > n, where n is the number of flowers, it's impossible to make m bouquets of k adjacent flowers. The function returns -1 in this case.
2. All flowers bloom on the same day: If all flowers bloom on the same day, the algorithm correctly identifies that the minimum number of days is that bloom day.
3. m or k is 1: If m or k is 1, the algorithm correctly identifies the minimum number of days is dependent on m <= n or k <= n.
4. Large bloomDay values: If bloomDay contains very large values, the binary search might take slightly longer, but the time complexity remains the same.