What is the time complexity of Maximum Total Beauty of the Gardens?

The optimal solution has a time complexity of O(n*log(target)*log(n)).

What data structures are used to solve Maximum Total Beauty of the Gardens?

This problem uses Array, Two Pointers, Binary Search, Greedy, Sorting, Enumeration, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Total Beauty of the Gardens?

Insight 1: **Greedy Approach for Full Gardens:** It's generally beneficial to try and make as many gardens 'full' as possible because the `full` multiplier contributes significantly to the total beauty. Therefore, start by considering how many gardens can be made complete, starting from the garden with the most flowers to reduce the cost to completion. Insight 2: **Binary Search for Optimal Minimum:** After determining the number of full gardens, we need to find the optimal minimum number of flowers among the remaining incomplete gardens. We can use binary search on the possible range of minimum flower counts (0 to `target - 1`) to find the best `best_level` efficiently.

Which companies ask LeetCode 2234. Maximum Total Beauty of the Gardens in interviews?

This problem is frequently asked at Intuit.

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Maximum Total Beauty of the Gardens - LeetCode 2234 Solution

Maximum Total Beauty of the Gardens - Complete Solution Guide

Maximum Total Beauty of the Gardens is LeetCode problem 2234, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to maximize the total beauty of `n` gardens by planting additional flowers. Each garden `i` initially has `flowers[i]` flowers. We can plant at most `newFlowers` additional flowers. A garden is considered complete if it has at least `target` flowers. The total beauty is calculated as `(number of complete gardens * full) + (minimum number of flowers in incomplete gardens * partial)`. If there are no incomplete gardens, the second term is 0. The goal is to return the maximum possible total beauty after planting the additional flowers optimally.

Solution Approach

The solution uses a combination of sorting, prefix sums, binary search, and a greedy strategy. First, it sorts the `flowers` array to easily determine which gardens are most cost-effective to complete. Then, it iterates through the possible number of complete gardens (`i`) from 0 to `n`. For each value of `i`, it calculates the cost (`cost_to_complete`) of making the last `i` gardens complete. If `cost_to_complete` exceeds `newFlowers`, it breaks the loop. Otherwise, it calculates the remaining flowers (`remaining_flowers`) and uses binary search to find the optimal minimum flower count (`best_level`) for the remaining incomplete gardens. Finally, it calculates the total beauty for the current configuration and updates the maximum beauty (`ans`) found so far.

Step-by-Step Algorithm

Step 1: **Sort the `flowers` array:** Sort the `flowers` array in ascending order. This makes it easier to prioritize gardens for completion based on their initial flower count.
Step 2: **Calculate Prefix Sums:** Calculate the prefix sum of the `flowers` array. This allows for efficient calculation of the cost to increase the flower count of incomplete gardens during the binary search.
Step 3: **Iterate through possible number of full gardens:** Loop through the possible number of full gardens, `i`, from 0 to `n`.
Step 4: **Calculate cost to complete `i` gardens:** For each `i`, calculate `cost_to_complete`, the total number of flowers needed to make the last `i` gardens complete (reaching the `target` value).
Step 5: **Check flower constraint:** If `cost_to_complete` exceeds `newFlowers`, break the loop because we cannot complete that many gardens.
Step 6: **Calculate remaining flowers:** If `cost_to_complete` is within the limit, calculate `remaining_flowers` as `newFlowers - cost_to_complete`.
Step 7: **Binary Search for optimal minimum:** Use binary search to find the optimal minimum flower count (`best_level`) for the remaining `n - i` incomplete gardens. The binary search searches the range [0, target - 1]. For each `mid` in this range, we calculate the cost to raise the flower count of all incomplete gardens to at least `mid`. The `best_level` will be the largest `mid` such that we can afford the cost.
Step 8: **Calculate and update maximum beauty:** Calculate the beauty using `i * full + best_level * partial` and update the `ans` with the maximum beauty found so far.
Step 9: **Return the maximum beauty:** After iterating through all possible numbers of full gardens, return the maximum beauty `ans`.

Key Insights

Insight 1: **Greedy Approach for Full Gardens:** It's generally beneficial to try and make as many gardens 'full' as possible because the `full` multiplier contributes significantly to the total beauty. Therefore, start by considering how many gardens can be made complete, starting from the garden with the most flowers to reduce the cost to completion.
Insight 2: **Binary Search for Optimal Minimum:** After determining the number of full gardens, we need to find the optimal minimum number of flowers among the remaining incomplete gardens. We can use binary search on the possible range of minimum flower counts (0 to `target - 1`) to find the best `best_level` efficiently.
Insight 3: **Prefix Sums for Cost Calculation:** Calculating the cost of increasing the flower count of incomplete gardens to a certain level involves summing the differences between the target level and the current flower count. Using prefix sums allows us to calculate this cost in O(1) time during the binary search, optimizing the overall time complexity.

Complexity Analysis

Time Complexity: O(n*log(target)*log(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Binary Search, Greedy, Sorting, Enumeration, Prefix Sum.

Companies

Asked at: Intuit.

Maximum Total Beauty of the Gardens | LeetCode 2234

Hard

Array Two Pointers Binary Search Greedy Sorting Enumeration Prefix Sum

Time: O(n*log(target)*log(n))Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long maximumBeauty(int[] flowers, long newFlowers, int target, int full, int partial) {
        int n = flowers.length;
        Arrays.sort(flowers);

        long[] prefix = new long[n + 1];
        for (int i = 0; i < n; i++) {
            prefix[i + 1] = prefix[i] + flowers[i];
        }

        long ans = 0;
        long cost_to_complete = 0;

        for (int i = 0; i <= n; i++) {
            if (i > 0) {
                long cost_one_more = Math.max(0, (long)target - flowers[n - i]);
                cost_to_complete += cost_one_more;
            }
            
            if (cost_to_complete > newFlowers) {
                break;
            }
            
            long remaining_flowers = newFlowers - cost_to_complete;
            
            int m = n - i;
            if (m == 0) {
                long beauty = (long)i * full;
                ans = Math.max(ans, beauty);
                continue;
            }

            int low = 0, high = target - 1;
            long best_level = 0;
            
            while (low <= high) {
                int mid = low + (high - low) / 2;
                
                int p_idx = binarySearch(flowers, 0, m, mid);
                
                long cost_partial = (long)p_idx * mid - prefix[p_idx];
                
                if (cost_partial <= remaining_flowers) {
                    best_level = mid;
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            
            long beauty = (long)i * full + best_level * partial;
            ans = Math.max(ans, beauty);
            }
        }
        
        return ans;
    }

    private int binarySearch(int[] arr, int low, int high, int target) {
        int l = low, r = high;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (arr[mid] < target) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        return l;
    }
}

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Problem Info

Description

Problem 2234: Maximum Total Beauty of the Gardens - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Greedy Sorting Enumeration Prefix Sum

Companies

Intuit

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2234

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: **Sort the `flowers` array:** Sort the `flowers` array in ascending order. This makes it easier to prioritize gardens for completion based on their initial flower count.
2
Step 2: **Calculate Prefix Sums:** Calculate the prefix sum of the `flowers` array. This allows for efficient calculation of the cost to increase the flower count of incomplete gardens during the binary search.
3
Step 3: **Iterate through possible number of full gardens:** Loop through the possible number of full gardens, `i`, from 0 to `n`.
4
Step 4: **Calculate cost to complete `i` gardens:** For each `i`, calculate `cost_to_complete`, the total number of flowers needed to make the last `i` gardens complete (reaching the `target` value).
5
Step 5: **Check flower constraint:** If `cost_to_complete` exceeds `newFlowers`, break the loop because we cannot complete that many gardens.
6
Step 6: **Calculate remaining flowers:** If `cost_to_complete` is within the limit, calculate `remaining_flowers` as `newFlowers - cost_to_complete`.
7
Step 7: **Binary Search for optimal minimum:** Use binary search to find the optimal minimum flower count (`best_level`) for the remaining `n - i` incomplete gardens. The binary search searches the range [0, target - 1]. For each `mid` in this range, we calculate the cost to raise the flower count of all incomplete gardens to at least `mid`. The `best_level` will be the largest `mid` such that we can afford the cost.
8
Step 8: **Calculate and update maximum beauty:** Calculate the beauty using `i * full + best_level * partial` and update the `ans` with the maximum beauty found so far.
9
Step 9: **Return the maximum beauty:** After iterating through all possible numbers of full gardens, return the maximum beauty `ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*log(target)*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The initial sorting takes O(n log n) time. The loop iterates `n` times. Inside the loop, the calculation of `cost_to_complete` takes O(n) overall because of the prefix sum. The binary search takes O(log target) time. Inside the binary search, we perform a bisect operation (either standard library or custom, depending on the language) which takes O(log n) time to determine the prefix index. The cost computation inside the binary search is constant, thus O(1). Overall, the time complexity is therefore dominated by the `n * log(target) * log(n)` factor from the outer loop and binary search.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Priority Queue (Heap)

Instead of iterating and using binary search, a priority queue could be used to keep track of the gardens requiring the least amount of additional flowers to complete. The flowers would be added to the complete set until `newFlowers` is exhausted. However, the heap doesn't readily provide the information to pick the optimal level for the incomplete gardens efficiently.

Time: O(n log n) for heap operations, potentially more complex for optimizationSpace: O(n) for the heap

Trade-offs: This approach might be useful if the number of gardens is extremely large, and only a small portion needs to be considered. However, it is harder to get right because figuring out the best level for partial gardens becomes more complex and computationally expensive.

✨

Key Insights

1Insight 1: **Greedy Approach for Full Gardens:** It's generally beneficial to try and make as many gardens 'full' as possible because the `full` multiplier contributes significantly to the total beauty. Therefore, start by considering how many gardens can be made complete, starting from the garden with the most flowers to reduce the cost to completion.
2Insight 2: **Binary Search for Optimal Minimum:** After determining the number of full gardens, we need to find the optimal minimum number of flowers among the remaining incomplete gardens. We can use binary search on the possible range of minimum flower counts (0 to `target - 1`) to find the best `best_level` efficiently.
3Insight 3: **Prefix Sums for Cost Calculation:** Calculating the cost of increasing the flower count of incomplete gardens to a certain level involves summing the differences between the target level and the current flower count. Using prefix sums allows us to calculate this cost in O(1) time during the binary search, optimizing the overall time complexity.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: **`newFlowers` is small:** If `newFlowers` is very small and we cannot complete any gardens, the binary search will optimize the minimum flower count for incomplete gardens.
Edge case 2: **`newFlowers` is very large:** If `newFlowers` is large enough to complete all gardens, the algorithm will consider this scenario and choose the optimal strategy (either make all gardens complete or some gardens complete and some incomplete).
Edge case 3: **`partial` is 0:** If `partial` is 0, the beauty is only determined by the number of full gardens, and the binary search will always choose 0. The main loop will still iterate through all possible numbers of full gardens to find the maximum.
Edge case 4: **`full` is 0:** If `full` is 0, the beauty is only determined by the minimum flower count in the incomplete gardens. The algorithm will still find the best value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: **Integer Overflow:** When dealing with large numbers like `newFlowers` and intermediate calculations, it's crucial to use `long long` or `long` (in Java) to prevent integer overflows. Failing to do so can lead to incorrect results.

✗Common mistake 2: **Incorrect Binary Search Implementation:** Implementing the binary search incorrectly can lead to suboptimal results. Ensure the binary search covers the entire range of possible minimum flower counts (0 to `target - 1`) and correctly calculates the cost of raising the flower count of incomplete gardens.

✗Common mistake 3: **Not sorting the flowers:** The solution relies on sorted data for efficient cost computation and binary search. Skipping the sorting step will result in incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: **Resource Allocation:** This problem is analogous to resource allocation scenarios where we have a limited budget (`newFlowers`) and want to maximize the overall benefit (`total beauty`) by allocating resources (`flowers`) to different projects (`gardens`).

🏦Real-world scenario 2: **Production Planning:** Imagine a manufacturing plant with different production lines. Each line needs a certain amount of resources to reach a target production level. We can use a similar approach to determine how to allocate a limited supply of resources to maximize the overall output and profit.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Total Beauty of the Gardens"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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