Solution Explanation: Minimum Number of Days to Make m Bouquets

This problem asks for the minimum number of days to make m bouquets, where each bouquet requires k adjacent blooming flowers. A binary search approach is highly efficient for this problem due to the monotonic nature of the solution: if you can make m bouquets by day x, you can certainly make them by day x+1.

Algorithm:

1. Binary Search Space: The search space is the range of possible days, from the minimum blooming day (1) to the maximum blooming day (let's call it max_day). We perform a binary search within this range.

2. Check Function (check(days)): The core of the solution lies in the check(days) function. This function takes a days value (potential answer) and determines if it's possible to make m bouquets by that day. It iterates through the bloomDay array:

   • If a flower blooms by days, it's added to the current bouquet count (cur).
   • If a flower doesn't bloom by days, the current bouquet count is reset to 0.
   • When cur reaches k (enough flowers for a bouquet), the bouquet count (cnt) is incremented, and cur is reset.
   • Finally, the function returns true if cnt >= m (enough bouquets are possible), otherwise false.

3. Binary Search Iteration: The binary search repeatedly halves the search space:

   • It calculates the middle day (mid).
   • It calls check(mid). If true, it means we might be able to achieve the goal with fewer days, so we search in the lower half (r = mid).
   • If false, we need more days, so we search in the upper half (l = mid + 1).

4. Result: The loop continues until l == r, at which point l (or r) represents the minimum number of days. If l is greater than max_day, it means it's impossible to make m bouquets, and -1 is returned.

Time Complexity: O(N log M), where N is the length of bloomDay and M is the maximum blooming day. The binary search takes O(log M) iterations, and each iteration involves a linear scan of bloomDay (O(N)) within the check function.

Space Complexity: O(1). The algorithm uses a constant amount of extra space.

Code Examples (Python and Java):

Python:

import bisect
 
class Solution:
    def minDays(self, bloomDay: List[int], m: int, k: int) -> int:
        def check(days: int) -> bool:
            cnt = cur = 0
            for x in bloomDay:
                cur = cur + 1 if x <= days else 0
                if cur == k:
                    cnt += 1
                    cur = 0
            return cnt >= m
 
        max_day = max(bloomDay)
        left = 1
        right = max_day + 1  # ensures that even the maximum day is checked.
 
        #Using bisect_left for efficient search instead of a while loop.
        result = bisect.bisect_left(range(right+1), True, key=check)
        return result if result <=max_day else -1
 

Java:

class Solution {
    public int minDays(int[] bloomDay, int m, int k) {
        int maxDay = 0;
        for (int day : bloomDay) maxDay = Math.max(maxDay, day);
 
        int left = 1, right = maxDay + 1; // to avoid off by one error
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (check(bloomDay, mid, m, k)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left > maxDay ? -1 : left; //if left is still greater than maxDay, it is impossible
    }
 
    private boolean check(int[] bloomDay, int days, int m, int k) {
        int bouquets = 0;
        int adjacentFlowers = 0;
        for (int day : bloomDay) {
            if (day <= days) {
                adjacentFlowers++;
                if (adjacentFlowers == k) {
                    bouquets++;
                    adjacentFlowers = 0;
                }
            } else {
                adjacentFlowers = 0;
            }
        }
        return bouquets >= m;
    }
}

These examples demonstrate the binary search and check function. Remember that other languages (C++, Go, TypeScript etc.) will have similar implementations, following the same algorithmic principles. The key is efficiently determining if a given number of days is sufficient to make the required bouquets using the check function.