BS-13. Minimum days to make M bouquets | Binary Search

Summary

This video walks through solving the “minimum days to make M bouquets” problem using binary search. It explains the problem with examples, defines the impossible case (when M × K > n), and shows how to check if a given day yields enough adjacent bloomed flowers to form bouquets. After a brute-force range-based approach, the instructor optimizes it with binary search over days (min bloom to max bloom), using a helper that counts consecutive bloomed flowers. The video concludes with implementation notes, complexity analysis (O(n · log(max−min))) and practical tips like preventing integer overflow and available code in multiple languages.

Takeaways

• 😀 The solution involves using binary search to find the optimal answer within a given range of values (from min to max).
• 😀 A function is used to determine the minimum and maximum values of the array, which are then used as bounds in the binary search.
• 😀 The binary search will check the feasibility of a solution by narrowing down the search space from high to low or vice versa based on conditions.
• 😀 If the number of tasks `n` is less than `M * K`, the problem is deemed impossible, and the algorithm should return `-1`.
• 😀 During binary search, if a solution is possible for a certain middle value, the algorithm eliminates the right half of the search space.
• 😀 If the solution is not possible with the current value, the algorithm increases the lower bound by eliminating the left half of the search space.
• 😀 A helper function checks whether a particular value (like `M` or `K`) can satisfy the problem's constraints.
• 😀 The time complexity of the approach is `O(n * log(max - min))`, where `n` is the size of the array, and `max` and `min` are the upper and lower bounds of the search space.
• 😀 The space complexity of the approach is constant (`O(1)`), as no extra space is used other than a few variables.
• 😀 To avoid integer overflow during multiplication (`M * K`), the algorithm recommends converting values to a larger integer type (like `long long`).
• 😀 The speaker encourages viewers to subscribe to the channel and follow on social media for more updates and tutorials.

Q & A

What is the main problem discussed in the video?

The video explains how to find the minimum number of days required to make M bouquets from N flowers, where each bouquet requires K adjacent bloomed flowers. The problem is solved using a binary search approach.

What does the 'BloomDay' array represent in the problem?

The 'BloomDay' array represents the day on which each flower blooms. For example, if BloomDay[i] = 7, the i-th flower will bloom on the 7th day.

What is the impossible case for this problem?

If the total number of flowers required (M × K) exceeds the number of available flowers (N), it is impossible to make the required bouquets. In that case, the function should return -1.

How is the 'possible' function designed to work?

The 'possible' function takes the BloomDay array, a day D, M, and K as inputs, and checks whether it is possible to make at least M bouquets by day D. It counts consecutive bloomed flowers and determines how many valid bouquets can be formed.

What is the brute-force approach to solve this problem?

The brute-force method iterates through each day between the minimum and maximum bloom days, checks whether it is possible to make the required bouquets on that day, and returns the smallest possible day. This approach has a time complexity of O(N × (max−min)).

Why is binary search applicable in this problem?

Binary search works because the solution space is monotonic — if it's possible to make M bouquets on day D, it will also be possible on any day greater than D. This property allows us to search efficiently for the minimum valid day.

What are the steps of the binary search approach in this problem?

1. Set low = min(BloomDay) and high = max(BloomDay). 2. Compute mid = (low + high) / 2. 3. Check if making M bouquets is possible on day mid. 4. If possible, move high to mid - 1 to find a smaller day; otherwise, move low to mid + 1. 5. The final answer is stored in 'low'.

What is the time and space complexity of the optimized binary search solution?

The time complexity is O(N × log(max−min)), where each binary search iteration takes O(N) to verify feasibility. The space complexity is O(1) because no additional data structures are used.

Why should the product of M and K be converted to a long data type in implementation?

Multiplying M and K directly might cause integer overflow in certain programming languages, especially when dealing with large inputs. Converting them to a long (or long long) ensures the calculation remains accurate.

What are the key takeaways from this problem-solving approach?

The key lessons include understanding how to identify impossible cases, developing a helper function to check feasibility, starting with a brute-force approach, and finally optimizing the solution using binary search based on monotonic behavior.

How does the 'counter' logic in the 'possible' function help determine bouquets?

The counter tracks consecutive bloomed flowers. Whenever a flower hasn't bloomed, the counter resets to zero. Each time the counter reaches K, it forms one bouquet and resets to zero to continue counting the next possible sequence.

What is the significance of finding the minimum and maximum bloom days before the search?

Determining the minimum and maximum bloom days defines the search boundaries for the binary search. It restricts the search space to only the range where a valid answer can exist, making the algorithm more efficient.