1482. Minimum Number of Days to Make m Bouquets LeetCode Solution
In this guide, you will get 1482. Minimum Number of Days to Make m Bouquets LeetCode Solution with the best time and space complexity. The solution to Minimum Number of Days to Make m Bouquets problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of 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 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 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.
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 <= 105
1 <= bloomDay[i] <= 109
1 <= m <= 106
1 <= k <= n
Complexity Analysis
Time Complexity: O(n\log (\max(\texttt{bloomDay})))
Space Complexity: O(1)
1482. Minimum Number of Days to Make m Bouquets LeetCode Solution in C++
class Solution {
public:
int minDays(vector<int>& bloomDay, int m, int k) {
if (bloomDay.size() < static_cast<long>(m) * k)
return -1;
int l = ranges::min(bloomDay);
int r = ranges::max(bloomDay);
while (l < r) {
const int mid = (l + r) / 2;
if (getBouquetCount(bloomDay, k, mid) >= m)
r = mid;
else
l = mid + 1;
}
return l;
}
private:
// Returns the number of bouquets (k flowers needed) can be made after the
// `waitingDays`..
int getBouquetCount(const vector<int>& bloomDay, int k, int waitingDays) {
int bouquetCount = 0;
int requiredFlowers = k;
for (const int day : bloomDay)
if (day > waitingDays) {
// Reset `requiredFlowers` since there was not enough adjacent flowers.
requiredFlowers = k;
} else if (--requiredFlowers == 0) {
// Use k adjacent flowers to make a bouquet.
++bouquetCount;
requiredFlowers = k;
}
return bouquetCount;
}
}; /* code provided by PROGIEZ */
1482. Minimum Number of Days to Make m Bouquets LeetCode Solution in Java
class Solution {
public int minDays(int[] bloomDay, int m, int k) {
if (bloomDay.length < (long) m * k)
return -1;
int l = Arrays.stream(bloomDay).min().getAsInt();
int r = Arrays.stream(bloomDay).max().getAsInt();
while (l < r) {
final int mid = (l + r) / 2;
if (getBouquetCount(bloomDay, k, mid) >= m)
r = mid;
else
l = mid + 1;
}
return l;
}
// Returns the number of bouquets (k flowers needed) can be made after the
// `waitingDays`..
private int getBouquetCount(int[] bloomDay, int k, int waitingDays) {
int bouquetCount = 0;
int requiredFlowers = k;
for (final int day : bloomDay)
if (day > waitingDays) {
// Reset `requiredFlowers` since there was not enough adjacent flowers.
requiredFlowers = k;
} else if (--requiredFlowers == 0) {
// Use k adjacent flowers to make a bouquet.
++bouquetCount;
requiredFlowers = k;
}
return bouquetCount;
}
} // code provided by PROGIEZ
1482. Minimum Number of Days to Make m Bouquets LeetCode Solution in Python
class Solution:
def minDays(self, bloomDay: list[int], m: int, k: int) -> int:
if len(bloomDay) < m * k:
return -1
def getBouquetCount(waitingDays: int) -> int:
"""
Returns the number of bouquets (k flowers needed) can be made after the
`waitingDays`.
"""
bouquetCount = 0
requiredFlowers = k
for day in bloomDay:
if day > waitingDays:
# Reset `requiredFlowers` since there was not enough adjacent flowers.
requiredFlowers = k
else:
requiredFlowers -= 1
if requiredFlowers == 0:
# Use k adjacent flowers to make a bouquet.
bouquetCount += 1
requiredFlowers = k
return bouquetCount
l = min(bloomDay)
r = max(bloomDay)
while l < r:
mid = (l + r) // 2
if getBouquetCount(mid) >= m:
r = mid
else:
l = mid + 1
return l # code by PROGIEZ
