Maximize the Beauty of the Garden

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ArraysGreedy AlgorithmsBinary Search

There is a garden of length n with n + 1 plots. The plots are numbered from 0 to n. In the i^th plot, there is a flower with beauty flowers[i].

You want to build k fences such that each fence can be placed between two adjacent plots. The cost of building a fence between the i^th and (i + 1)^th plots is fenceCost[i]. The total cost of building all fences should not exceed maxCost.

A fence divides the garden into sections. The beauty of each section is the sum of the beauties of all the flowers in that section. The beauty of the garden is the sum of the beauties of all the sections.

Return the maximum beauty of the garden after building all the fences.

Example 1:

Input: flowers = [1,2,3,1,2], fenceCost = [1,2,3,1], k = 2, maxCost = 5
Output: 14
Explanation:
Place fences between plot 0 and 1, and between plot 3 and 4.
The sections are divided as: [1], [2,3,1], [2].
The beauty of the garden is: 1 + (2+3+1) + 2 = 1 + 6 + 2 = 9.
It can be proved that you cannot get a higher beauty with any other fences.

Example 2:

Input: flowers = [5,6,1,4,2], fenceCost = [1,3,2,5], k = 3, maxCost = 9
Output: 19
Explanation:
Place fences between plot 0 and 1, between plot 2 and 3, and between plot 3 and 4.
The sections are divided as: [5], [6,1], [4], [2].
The beauty of the garden is: 5 + (6+1) + 4 + 2 = 5 + 7 + 4 + 2 = 18.

Example 3:

Input: flowers = [1,2,3,4,5,6], fenceCost = [6,4,3,2,1], k = 1, maxCost = 8
Output: 21
Explanation:
Place fence between plot 4 and 5.
The sections are divided as: [1,2,3,4,5], [6].
The beauty of the garden is: (1+2+3+4+5) + 6 = 15 + 6 = 21.

Constraints:

3 <= n <= 10⁵
flowers.length == n + 1
fenceCost.length == n
1 <= flowers[i] <= 10⁹
1 <= fenceCost[i] <= 10⁹
1 <= k <= n / 2
k * min(fenceCost[i]) <= maxCost <= 10¹⁵

Try coding it in LeetCode

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 possible ranges for the beauty values of each flower and the number of flowers allowed in each plot?
Can the number of flowers I'm allowed to plant in each plot be zero?
If it's impossible to achieve a positive total beauty, what should I return?
Are there any restrictions on the total number of plots available, or is it effectively unlimited?
Can the beauty values of the flowers be negative?

Brute Force Solution

Approach

The brute force approach to maximizing garden beauty involves checking absolutely every possible way to plant flowers in different locations. We consider every conceivable arrangement, calculating the beauty for each one. Finally, we pick the arrangement that yields the highest beauty score.

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

Consider planting flowers in the very first location.
Now, consider not planting flowers in the very first location.
For each of these choices, repeat the same decision-making process for the second location: plant or don't plant.
Continue this process for every location in the garden.
For each complete arrangement of flowers you've created, calculate the total 'beauty' of the garden based on how the flowers are arranged relative to each other and other features.
Compare the beauty of all possible arrangements.
Select the arrangement that results in the maximum total beauty score; that's the best planting arrangement.

Code Implementation

def maximize_garden_beauty_brute_force(garden_locations, beauty_function):
    max_beauty = float('-inf')

    def calculate_beauty(flower_arrangement):
        return beauty_function(flower_arrangement)

    def generate_arrangements(index, current_arrangement):
        nonlocal max_beauty
        # If we've reached the end of the garden, calculate the beauty
        if index == len(garden_locations):
            garden_beauty = calculate_beauty(current_arrangement)
            max_beauty = max(max_beauty, garden_beauty)
            return

        # Explore the option of not planting a flower at this location

        generate_arrangements(index + 1, current_arrangement + [0])

        # Explore the option of planting a flower at this location
        # We are trying all possibilities by either planting or not planting at each step
        generate_arrangements(index + 1, current_arrangement + [1])

    # Initiate recursive function
    generate_arrangements(0, [])

    return max_beauty

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores every possible subset of garden locations. For each of the 'n' locations in the garden, we have two choices: plant a flower or don't plant a flower. This leads to 2 * 2 * ... * 2 (n times) possible combinations or arrangements. Therefore, the time complexity is O(2^n) as we evaluate each of these combinations to find the maximum beauty.

Space Complexity

O(N) – The brute force approach described explores every possible arrangement using recursion. Each level of recursion corresponds to a location in the garden. The maximum depth of the recursion is N, where N is the number of locations in the garden. Each recursive call requires a stack frame, so the maximum space used by the call stack is proportional to N, leading to O(N) auxiliary space.

Optimal Solution

Approach

The best way to maximize the garden's beauty is to strategically decide how many flowers to plant in each flowerbed. We want to favor complete flowerbeds while minimizing leftover soil that doesn't fill a bed.

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

First, figure out the total beauty you can achieve if you don't plant any new flowers at all.
Then, imagine you have infinite new flowers. Calculate the maximum beauty you could have if every flowerbed was full.
Now, consider all the possibilities between these two extremes. For each option, figure out how many new flowers you'll need to buy to reach it, and how many leftover flowers will remain.
Start from the option where no flowers are purchased, and keep track of the most beauty you can have while still being able to afford the new flowers.
As you examine possibilities that are further away from the initial state, if it exceeds our max flowers, stop pursuing that state.
The final answer is the largest value for the garden's beauty that can be reached given your flower budget.

Code Implementation

def maximize_the_beauty_of_the_garden(flowerbeds, new_flowers, target, full, partial):
    number_of_flowerbeds = len(flowerbeds)

    initial_beauty = 0
    for flowers in flowerbeds:
        initial_beauty += min(flowers, target) * partial
        if flowers >= target:
            initial_beauty += full

    max_possible_beauty = number_of_flowerbeds * full + sum(target * partial for _ in range(number_of_flowerbeds))
    flowerbeds.sort()

    max_beauty_achieved = initial_beauty

    for flowerbeds_filled in range(number_of_flowerbeds + 1):
        flowers_needed = 0
        current_beauty = 0

        # Calculate flowers needed and beauty for filling flowerbeds.
        for i in range(flowerbeds_filled):
            flowers_needed += max(0, target - flowerbeds[number_of_flowerbeds - 1 - i])
            current_beauty += full

        if flowers_needed > new_flowers:
            break

        # If there are flowers still left, add partial beauty.
        remaining_flowers = new_flowers - flowers_needed
        
        for i in range(number_of_flowerbeds - flowerbeds_filled):
            current_beauty += min(flowerbeds[i] + remaining_flowers, target) * partial
            if flowerbeds[i] + remaining_flowers >= target:
                current_beauty += full
        else:
            # This is necessary to handle edge cases
            # Where remaining flowers are not enough to fill the flowerbeds
            max_beauty_achieved = max(max_beauty_achieved, current_beauty)
            continue

        max_beauty_achieved = max(max_beauty_achieved, current_beauty)

    return max_beauty_achieved

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through all possible levels of flowerbed completion. In the worst case, this involves iterating through all n flowerbeds. Within each iteration, it calculates the number of flowers needed, involving sorting the flowerbeds based on their current flower count, which takes O(n log n) time. Therefore, the overall time complexity is dominated by the sorting step within each of the n iterations, resulting in O(n log n).

Space Complexity

O(1) – The provided plain English explanation primarily describes iterative calculations and comparisons. It mentions tracking the maximum beauty and the number of flowers needed, suggesting a few variables to store these values. These variables consume a constant amount of space regardless of the input size N (number of flowerbeds), therefore the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty gardens array	Return 0 immediately, as there's no garden to beautify.
Capacity is zero	Return 0 immediately as no flowers can be planted to increase beauty.
All gardens have zero beauty	Plant flowers in the garden with the lowest cost to maximize the overall beauty.
Cost array is empty	Return 0 if gardens array is also empty; otherwise throw an IllegalArgumentException.
Integer overflow when calculating beauty	Use long data type to store intermediate beauty values and compare with the maximum value.
Capacity is extremely large, but number of gardens is small	Plant flowers in the gardens proportionally to the cost until the capacity is exhausted.
Cost array contains very large or very small numbers	Ensure multiplication or division involving costs does not lead to overflow or underflow, using long if needed.
No flowers can be planted to achieve positive beauty with a given cost or arrangement	Return the beauty of the garden without planting any flowers.

Empty gardens array

How to Handle:

Return 0 immediately, as there's no garden to beautify.

Capacity is zero

How to Handle:

Return 0 immediately as no flowers can be planted to increase beauty.

All gardens have zero beauty

How to Handle:

Plant flowers in the garden with the lowest cost to maximize the overall beauty.

Cost array is empty

How to Handle:

Return 0 if gardens array is also empty; otherwise throw an IllegalArgumentException.

Integer overflow when calculating beauty

How to Handle:

Use long data type to store intermediate beauty values and compare with the maximum value.

Capacity is extremely large, but number of gardens is small

How to Handle:

Plant flowers in the gardens proportionally to the cost until the capacity is exhausted.

Cost array contains very large or very small numbers

How to Handle:

Ensure multiplication or division involving costs does not lead to overflow or underflow, using long if needed.

No flowers can be planted to achieve positive beauty with a given cost or arrangement

How to Handle:

Return the beauty of the garden without planting any flowers.