Minimum Number of Days to Make m Bouquets

Problem Statement:

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 ith 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 mbouquets 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.

Constraints: 😈

• bloomDay.length == n
• 1 <= n <= 10^5
• 1 <= bloomDay[i] <= 10^9
• 1 <= m <= 10^6
• 1 <= k <= n

Solution:

• Time Complexity : $O(n\cdot log(maxBloomDay))$

class Solution {
 
    // Helper method to check if it's possible to make 'm' bouquets in 'days' time
    private boolean feasible(int days, int m, int k, int[] bloomDay) {
        int flower = 0; // Counter for consecutive flowers in bloom
        int bouquet = 0; // Counter for bouquets made
 
        // Iterate over each bloom day
        for (int bloom : bloomDay) {
            if (days >= bloom) { // Flower is in bloom within 'days'
                flower += 1; // Increment the count of blooming flowers
                if (flower == k) { // Check if we have enough flowers for a bouquet
                    bouquet += 1; // Make one bouquet
                    flower = 0; // Reset flower count for the next bouquet
                }
            } else {
                flower = 0; // Reset flower count if a non-blooming day is found
            }
        }
        return bouquet >= m; // Return true if required bouquets can be made
    }
 
    public int minDays(int[] bloomDay, int m, int k) {
	    // Check if there are enough flowers to make 'm' bouquets
        if ((long) bloomDay.length < (long) m * k) {
            return -1; // Return -1 if it's not possible to make the bouquets
        }
 
		// *** Define the search space *** 
        int maxBloomDay = Integer.MIN_VALUE; // Initialize max bloom day to minimum integer value
 
        // Find the maximum bloom day
        for (int day : bloomDay) {
            maxBloomDay = Math.max(maxBloomDay, day);
        }
		
		
        int left = 1; // Minimum possible day
        int right = maxBloomDay; // Maximum possible day
 
        // Binary search to find the minimum day required
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (feasible(mid, m, k, bloomDay)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
 
        return left;
    }
}