1482. Minimum Number of Days to Make m Bouquets #

Problem #

Given an integer array bloomDay, an integer m and an integer k.

We need 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 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's see what happened in the first three days. x means flower bloomed and _ means flower didn't 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's 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.

Example 4:

Input: bloomDay = [1000000000,1000000000], m = 1, k = 1
Output: 1000000000
Explanation: You need to wait 1000000000 days to have a flower ready for a bouquet.

Example 5:

Input: bloomDay = [1,10,2,9,3,8,4,7,5,6], m = 4, k = 2
Output: 9

Constraints:

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

Problem Summary #

Given an integer array bloomDay, and two integers m and k. You need to make m bouquets. To make a bouquet, you need to use k adjacent flowers from the garden. There are n flowers in the garden; the ith flower will bloom on bloomDay[i], and can be used in exactly one bouquet. Return the minimum number of days you need to wait to pick m bouquets from the garden. If it is impossible to pick m bouquets, return -1.

Solution Approach #

This problem is a binary search problem. The solution space is fixed, and the answer interval must be within [0, maxDay]. This is a monotonically increasing and ordered interval, so binary search can be used within this solution space. Find the first solution in the interval [0, maxDay] that can satisfy m bouquets. The condition for binary search to determine whether it is true is: traverse the array from left to right, and count whether the flowers have bloomed under the current date. If k flowers bloom consecutively, that makes 1 bouquet. After the array traversal ends, if the total number of bouquets ≥ k, then it is the answer. Binary search will return the smallest index, which corresponds to the minimum number of days that satisfies the problem requirement.

Code #

package leetcode

import "sort"

func minDays(bloomDay []int, m int, k int) int {
	if m*k > len(bloomDay) {
		return -1
	}
	maxDay := 0
	for _, day := range bloomDay {
		if day > maxDay {
			maxDay = day
		}
	}
	return sort.Search(maxDay, func(days int) bool {
		flowers, bouquets := 0, 0
		for _, d := range bloomDay {
			if d > days {
				flowers = 0
			} else {
				flowers++
				if flowers == k {
					bouquets++
					flowers = 0
				}
			}
		}
		return bouquets >= m
	})
}