Maximum Total Beauty of the Gardens

Hard

Intuit

1 view

Topics:

ArraysBinary SearchGreedy Algorithms

Alice is a caretaker of n gardens and she wants to plant flowers to maximize the total beauty of all her gardens.

You are given a 0-indexed integer array flowers of size n, where flowers[i] is the number of flowers already planted in the i^th garden. Flowers that are already planted cannot be removed. You are then given another integer newFlowers, which is the maximum number of flowers that Alice can additionally plant. You are also given the integers target, full, and partial.

A garden is considered complete if it has at least target flowers. The total beauty of the gardens is then determined as the sum of the following:

The number of complete gardens multiplied by full.
The minimum number of flowers in any of the incomplete gardens multiplied by partial. If there are no incomplete gardens, then this value will be 0.

Return the maximum total beauty that Alice can obtain after planting at most newFlowers flowers.

Example 1:

Input: flowers = [1,3,1,1], newFlowers = 7, target = 6, full = 12, partial = 1
Output: 14
Explanation: Alice can plant
- 2 flowers in the 0^th garden
- 3 flowers in the 1^st garden
- 1 flower in the 2^nd garden
- 1 flower in the 3^rd garden
The gardens will then be [3,6,2,2]. She planted a total of 2 + 3 + 1 + 1 = 7 flowers.
There is 1 garden that is complete.
The minimum number of flowers in the incomplete gardens is 2.
Thus, the total beauty is 1 * 12 + 2 * 1 = 12 + 2 = 14.
No other way of planting flowers can obtain a total beauty higher than 14.

Example 2:

Input: flowers = [2,4,5,3], newFlowers = 10, target = 5, full = 2, partial = 6
Output: 30
Explanation: Alice can plant
- 3 flowers in the 0^th garden
- 0 flowers in the 1^st garden
- 0 flowers in the 2^nd garden
- 2 flowers in the 3^rd garden
The gardens will then be [5,4,5,5]. She planted a total of 3 + 0 + 0 + 2 = 5 flowers.
There are 3 gardens that are complete.
The minimum number of flowers in the incomplete gardens is 4.
Thus, the total beauty is 3 * 2 + 4 * 6 = 6 + 24 = 30.
No other way of planting flowers can obtain a total beauty higher than 30.
Note that Alice could make all the gardens complete but in this case, she would obtain a lower total beauty.

Constraints:

1 <= flowers.length <= 10⁵
1 <= flowers[i], target <= 10⁵
1 <= newFlowers <= 10¹⁰
1 <= full, partial <= 10⁵

Maximum Total Beauty of the Gardens

Hard

Intuit

1 view

Topics:

ArraysBinary SearchGreedy Algorithms

Alice is a caretaker of n gardens and she wants to plant flowers to maximize the total beauty of all her gardens.

You are given a 0-indexed integer array flowers of size n, where flowers[i] is the number of flowers already planted in the i^th garden. Flowers that are already planted cannot be removed. You are then given another integer newFlowers, which is the maximum number of flowers that Alice can additionally plant. You are also given the integers target, full, and partial.

A garden is considered complete if it has at least target flowers. The total beauty of the gardens is then determined as the sum of the following:

The number of complete gardens multiplied by full.
The minimum number of flowers in any of the incomplete gardens multiplied by partial. If there are no incomplete gardens, then this value will be 0.

Return the maximum total beauty that Alice can obtain after planting at most newFlowers flowers.

Example 1:

Input: flowers = [1,3,1,1], newFlowers = 7, target = 6, full = 12, partial = 1
Output: 14
Explanation: Alice can plant
- 2 flowers in the 0^th garden
- 3 flowers in the 1^st garden
- 1 flower in the 2^nd garden
- 1 flower in the 3^rd garden
The gardens will then be [3,6,2,2]. She planted a total of 2 + 3 + 1 + 1 = 7 flowers.
There is 1 garden that is complete.
The minimum number of flowers in the incomplete gardens is 2.
Thus, the total beauty is 1 * 12 + 2 * 1 = 12 + 2 = 14.
No other way of planting flowers can obtain a total beauty higher than 14.

Example 2:

Input: flowers = [2,4,5,3], newFlowers = 10, target = 5, full = 2, partial = 6
Output: 30
Explanation: Alice can plant
- 3 flowers in the 0^th garden
- 0 flowers in the 1^st garden
- 0 flowers in the 2^nd garden
- 2 flowers in the 3^rd garden
The gardens will then be [5,4,5,5]. She planted a total of 3 + 0 + 0 + 2 = 5 flowers.
There are 3 gardens that are complete.
The minimum number of flowers in the incomplete gardens is 4.
Thus, the total beauty is 3 * 2 + 4 * 6 = 6 + 24 = 30.
No other way of planting flowers can obtain a total beauty higher than 30.
Note that Alice could make all the gardens complete but in this case, she would obtain a lower total beauty.

Constraints:

1 <= flowers.length <= 10⁵
1 <= flowers[i], target <= 10⁵
1 <= newFlowers <= 10¹⁰
1 <= full, partial <= 10⁵

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the input values such as `flowers[i]`, `newFlowers`, `target`, `full`, and `partial`? Specifically, what are the maximum and minimum possible values, and are negative values allowed?
If it's impossible to make even a single garden complete (i.e., reach `target` flowers in at least one garden, even after using all `newFlowers`), what should the function return? Should I optimize for partial gardens in that case?
If there are multiple ways to achieve the maximum total beauty, is any one valid solution acceptable, or is there a specific criterion for selecting among them?
If `partial` is zero, does that mean I don't need to consider the beauty from incomplete gardens, and can optimize just for the number of complete gardens?
Can `flowers` be empty or null? If so, what should I return?

Brute Force Solution

Approach

The brute force method solves this garden beauty problem by trying every single possible combination of how much to grow each garden. We will examine each of these possibilities to identify the arrangement that gives us the highest possible total beauty.

Here's how the algorithm would work step-by-step:

Start by considering planting the smallest possible amount (possibly zero) in each garden.
Calculate the total beauty of the gardens for this planting arrangement.
Now, systematically increase the planting in the first garden by one, and calculate the new total beauty. Then, increase it again and so on, checking the total beauty each time. Repeat this process until the planting amount for the first garden reaches a maximum limit.
After exhausting all planting amounts for the first garden, reset it to its smallest value (possibly zero). Then, increment the planting amount for the second garden by one. Now, repeat the process from step three, cycling through every planting amount in the first garden for this new planting amount in the second garden.
Continue this pattern for all gardens. Each time you consider new planting amounts, make sure you haven't exceeded the total amount of planting you're allowed to do.
Keep track of the total beauty for every possible valid planting arrangement that you have encountered.
Finally, once you've tried all combinations, choose the planting arrangement that resulted in the highest total beauty. That's your answer!

Code Implementation

def maximum_total_beauty_brute_force(flowers, new_flowers, target, full, partials):
    number_of_gardens = len(flowers)
    maximum_beauty = 0

    def calculate_beauty(flower_allocation):
        total_beauty = 0
        full_gardens = 0
        for flower_count in flower_allocation:
            if flower_count >= target:
                total_beauty += full
                full_gardens += 1
            else:
                total_beauty += partials[flower_count]
        return total_beauty

    def distribute_flowers(garden_index, remaining_new_flowers, flower_allocation):
        nonlocal maximum_beauty

        if garden_index == number_of_gardens:
            # All gardens have been assigned flowers, calculate total beauty.
            total_beauty = calculate_beauty(flower_allocation)
            maximum_beauty = max(maximum_beauty, total_beauty)
            return

        # Try all possible flower allocations for the current garden.
        max_flowers_for_garden = flowers[garden_index] + remaining_new_flowers
        for flowers_to_add in range(remaining_new_flowers + 1):
            current_garden_flowers = flowers[garden_index] + flowers_to_add

            # Recursive step: move to the next garden.
            distribute_flowers(
                garden_index + 1,
                remaining_new_flowers - flowers_to_add,
                flower_allocation + [current_garden_flowers],
            )

    # Start the flower distribution process from the first garden
    distribute_flowers(0, new_flowers, [])

    return maximum_beauty

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(m^n) – Let n be the number of gardens and m be the maximum possible planting amount for each garden (or a reasonable upper bound if unlimited). The brute force approach iterates through all possible planting amounts for each of the n gardens. If we assume each garden can have up to m different planting amounts, the algorithm explores m possibilities for the first garden, m for the second, and so on, up to the nth garden. Therefore, the total number of combinations explored is m * m * ... * m (n times), which equals m^n. Hence, the time complexity is O(m^n).

Space Complexity

O(1) – The brute force solution, as described, does not use any auxiliary data structures that scale with the input size. It only requires a few variables to keep track of the current planting arrangement being considered, the best total beauty found so far, and loop counters for iterating through all possible combinations. The space used by these variables remains constant regardless of the number of gardens or the total amount of planting allowed, therefore the space complexity is O(1).

Optimal Solution

Approach

The problem asks us to maximize beauty by strategically upgrading gardens. A key insight is that we can use binary search to efficiently find the optimal beauty threshold. We try to achieve a target beauty and adjust our strategy based on whether it's achievable.

Here's how the algorithm would work step-by-step:

First, recognize that the minimum possible beauty and the maximum possible beauty can be calculated. Our target beauty will lie between these two values.
Use binary search to find the optimal target beauty. The binary search will give us a potential 'target beauty' value to try.
For a given target beauty, check if it's possible to achieve that level of beauty in all gardens. This involves calculating how much we need to invest in each garden to reach the target beauty.
Sum up the total investment needed to reach the target beauty across all gardens.
Compare the total investment with the available investment. If the total investment is less than or equal to the available investment, then it's feasible to achieve that beauty.
If the target beauty is feasible, it means we can potentially aim for an even higher beauty. So, adjust the binary search to look for a higher target.
If the target beauty is not feasible, it means we need to aim for a lower beauty. So, adjust the binary search to look for a lower target.
During the binary search, keep track of the highest feasible target beauty you find. This will be our answer.
Finally, after the binary search is done, calculate the precise total beauty achieved. This involves figuring out how much investment is leftover and distributing it optimally among the gardens. This part makes sure we're using every last bit of investment for extra beauty.

Code Implementation

def maximum_beauty(flowers, new_flowers, target, full, partial):
    number_of_gardens = len(flowers)
    flowers.sort()
    prefix_sum = [0] * (number_of_gardens + 1)
    for i in range(number_of_gardens):
        prefix_sum[i + 1] = prefix_sum[i] + flowers[i]

    max_total_beauty = 0
    for complete_gardens in range(number_of_gardens + 1):
        # Iterate to find optimal number of fully beautified gardens.
        remaining_flowers = new_flowers
        current_total_beauty = 0

        # Attempt to make 'complete_gardens' gardens fully beautiful
        flowers_needed = 0
        for i in range(number_of_gardens - complete_gardens, number_of_gardens):
            flowers_needed += max(0, target - flowers[i])

        if flowers_needed <= remaining_flowers:
            current_total_beauty += complete_gardens * full
            remaining_flowers -= flowers_needed
            remaining_gardens = number_of_gardens - complete_gardens

            # Binary search to maximize beauty from partial gardens.
            if remaining_gardens > 0:
                low = 0
                high = target - 1
                best_base_beauty = 0

                while low <= high:
                    mid = (low + high) // 2
                    required_flowers = remaining_gardens * mid - (prefix_sum[number_of_gardens - complete_gardens] - prefix_sum[0])

                    if required_flowers <= remaining_flowers:
                        best_base_beauty = mid
                        low = mid + 1
                    else:
                        high = mid - 1

                current_total_beauty += best_base_beauty * partial
            max_total_beauty = max(max_total_beauty, current_total_beauty)
    return max_total_beauty

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log(maxBeauty)) – The algorithm utilizes binary search to find the optimal target beauty. The binary search runs in O(log(maxBeauty)) where maxBeauty is the maximum possible beauty, determined by the gardens array. Inside the binary search loop, we iterate through all the gardens to calculate the required investment. This iteration takes O(n) time, where n is the number of gardens. Therefore, the overall time complexity is O(n log(maxBeauty)).

Space Complexity

O(1) – The algorithm primarily uses a binary search approach. It involves calculating target beauty values and investment amounts, but these are stored in a few constant space variables. No auxiliary data structures like arrays or hash maps are created that scale with the input size, N (the number of gardens). Therefore, the auxiliary space complexity remains constant, independent of the input.

Edge Cases

Case	How to Handle
Empty flowers array	If flowers is empty, there are no gardens; return 0 since no beauty can be created.
newFlowers is 0	If newFlowers is 0, no flowers can be planted; calculate the beauty based on initial flower counts.
All flowers are already >= target	If all flowers are already at or above the target, maximize the beauty using full * n.
target is very large relative to flowers and newFlowers	If target is significantly larger than flowers and newFlowers combined, the optimal strategy might be to focus all newFlowers on one garden.
full and partial are 0	If full and partial are zero, the total beauty will always be zero regardless of flower distribution, so return 0.
Integer overflow when calculating beauty	Use long type to store intermediate and final beauty values to prevent integer overflow.
newFlowers can make all gardens complete	If newFlowers is large enough to complete all gardens, plant until all gardens have at least target flowers.
duplicate flower counts in the input array	The solution needs to handle duplicate flower counts correctly during optimization, ensuring not to double-count beauty.