Leetcode #1482: Minimum Number of Days to Make m Bouquets
In this guide, we solve Leetcode #1482 Minimum Number of Days to Make m Bouquets in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer array bloomDay, an integer m and an integer k. You want to make m bouquets.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Binary Search
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
class Solution:
def minDays(self, bloomDay: List[int], m: int, k: int) -> int:
def check(days: int) -> int:
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
mx = max(bloomDay)
l = bisect_left(range(mx + 2), True, key=check)
return -1 if l > mx else l
Complexity
The time complexity is , where and are the number of flowers in the garden and the maximum blooming day, respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.