[Array] Can Place Flowers - Easy

The “Can Place Flowers” problem is a well-known challenge in algorithms. It asks if we can plant a certain number of flowers in a flowerbed. We must remember that no two flowers can sit next to each other. To solve this, we usually look through the flowerbed array. We check the spots to see if we can plant flowers while keeping the nearby spaces empty.

In this article, we will look at different ways to solve the “Can Place Flowers” problem. First, we will explain the problem. Then, we will see both brute force and better greedy algorithm solutions in Java, Python, and C++. At the end, we will check the time complexity of these solutions. We will also answer some common questions.

• [Array] Can Place Flowers - Easy Solution Overview
  
• Understanding the Problem Statement for Array Can Place Flowers
  
• Brute Force Approach in Java for Can Place Flowers
  
• Optimized Approach Using Greedy Algorithm in Java
  
• Python Implementation of Brute Force for Can Place Flowers
  
• Optimized Greedy Solution in Python for Can Place Flowers
  
• C++ Brute Force Method for Can Place Flowers
  
• C++ Optimized Greedy Algorithm for Can Place Flowers
  
• Time Complexity Analysis of Array Can Place Flowers Solutions
  
• Frequently Asked Questions

If you want to learn more about similar topics, you can check these articles: Array Two Sum - Easy, Array Best Time to Buy and Sell Stock - Easy, and Array Contains Duplicate - Easy.

Understanding the Problem Statement for Array Can Place Flowers

The problem “Can Place Flowers” is about finding out if we can plant a certain number of flowers in a flowerbed. We need to follow the rule of no adjacent flowers. We have an array called flowerbed where:

• flowerbed[i] is 0 if the i-th spot is empty and 1 if it has a flower.
• We can only plant a flower in an empty spot (0) when the neighboring spots are also empty (0).

Our job is to check if we can plant n new flowers in the flowerbed without breaking the rule.

Input:

• An array flowerbed made of integers (0s and 1s).
• An integer n shows how many flowers we want to plant.

Output:

• We return true if we can plant n new flowers in the flowerbed without breaking the no-adjacent-flowers rule. If we can’t, we return false.

Example:

Input: flowerbed = [1,0,0,0,1], n = 1
Output: true

Input: flowerbed = [1,0,0,0,1], n = 2
Output: false

Constraints:

• The length of the flowerbed array will be from 1 to 20000.
• The flowerbed will only have 0s and 1s.

We can solve this problem by using a simple method or by using a greedy approach. This can help us check the possible spots for the flowers more efficiently.

For more information on similar array problems, we can visit Array Two Sum and Array Best Time to Buy and Sell Stock.

Brute Force Approach in Java for Can Place Flowers

We can use the brute force approach for the “Can Place Flowers” problem. This means checking every spot in the flowerbed to see if we can plant a flower without breaking the rule of not having flowers next to each other. We will go through the flowerbed array, checking each position. We make sure that planting a flower there will not cause adjacent flowers.

Java Implementation

public class Flowerbed {
    public static boolean canPlaceFlowers(int[] flowerbed, int n) {
        int count = 0;
        for (int i = 0; i < flowerbed.length; i++) {
            // Check if the current spot is empty and the spots next to it are empty or out of bounds
            if (flowerbed[i] == 0 && 
                (i == 0 || flowerbed[i - 1] == 0) && 
                (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) {
                flowerbed[i] = 1; // Plant a flower
                count++;
            }
            if (count >= n) {
                return true;
            }
        }
        return count >= n;
    }

    public static void main(String[] args) {
        int[] flowerbed = {1, 0, 0, 0, 1};
        int n = 1;
        System.out.println(canPlaceFlowers(flowerbed, n)); // Output: true
    }
}

Explanation

• Initialization: We start with a counter called count. This will keep track of how many flowers we have planted.
• Loop through flowerbed: For each spot in the flowerbed:
  • We check if the current spot is empty (flowerbed[i] == 0).
  • We also check if the left and right spots are empty or out of bounds.
• Plant a flower: If all conditions are good, we plant a flower by setting flowerbed[i] to 1 and increase count.
• Check count: If count gets to n, we return true.
• Final check: After the loop, we check if the total number of planted flowers is at least n.

This brute force method has time complexity of O(m), where m is the size of the flowerbed array. It is easy to understand but might not be the best solution for bigger inputs.

For more related problems with arrays, we can check out Array Two Sum and Array Best Time to Buy and Sell Stock.

Optimized Approach Using Greedy Algorithm in Java

We can solve the “Can Place Flowers” problem with a better way. We use a greedy algorithm. This method helps us find the most flowers we can plant in a flowerbed. The greedy strategy looks at each spot in the flowerbed and checks if we can plant a flower without breaking the rules.

Algorithm Steps:

1. Initialization: First, we need a counter. This counter will count the flowers we plant. We also set an index to go through the flowerbed.
2. Traversal: We loop through the flowerbed array and check:
   • If the current spot is empty (0).
   • If the spot before it is empty or if we are at the start of the array.
   • If the spot after it is empty or if we are at the end of the array.
3. Placement: If all checks pass, we place a flower (change the current spot to 1) and add one to the counter.
4. Skip: We then move to the position after where we just planted the flower. This helps keep the distance rule.

Java Implementation:

We can write the greedy algorithm in Java like this:

public class Flowerbed {
    public static int maxFlowers(int[] flowerbed, int n) {
        int count = 0;
        int i = 0;
        
        while (i < flowerbed.length) {
            if (flowerbed[i] == 0 && 
                (i == 0 || flowerbed[i - 1] == 0) && 
                (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) {
                
                flowerbed[i] = 1; // Place a flower
                count++; // Increment the count of flowers
            }
            i++; // Move to the next position
        }
        
        return count >= n; // Return if we can plant at least n flowers
    }

    public static void main(String[] args) {
        int[] flowerbed = {1, 0, 0, 0, 1};
        int n = 1;
        boolean result = maxFlowers(flowerbed, n);
        System.out.println(result); // Output: true
    }
}

Explanation of Code:

• The function maxFlowers checks each position in the flowerbed.
• It makes sure we can plant a flower only if the current spot, the one before, and the one after are all suitable (this means no flowers can be next to each other).
• The result gives a boolean value. It tells us if we can plant at least n flowers.

This greedy solution works fast. It has a time complexity of O(m). Here, m is the size of the flowerbed. This makes it good for big inputs. If you want to learn more about how to handle arrays, you can check articles like Array Contains Duplicate.

Python Implementation of Brute Force for Can Place Flowers

The brute force way to solve the “Can Place Flowers” problem is to look at every spot in the flowerbed. We check if we can plant a flower. The rules say that no two flowers can be next to each other and a flower can only go in an empty spot.

Problem Definition

We have an integer array flowerbed. In this array, 0 means the spot is empty and 1 means a flower is already planted. We also have an integer n which shows how many flowers we want to plant. Our goal is to see if we can plant n new flowers in the flowerbed without breaking the rule of no-adjacent-flowers.

Brute Force Approach Implementation

The brute force method looks at each spot in the flowerbed array. If a spot is empty and the spots next to it are also empty or out of bounds, we can plant a flower there. We keep going until we either plant all flowers or we find out that we cannot plant enough flowers.

Here is a simple Python code for the brute force method:

def canPlaceFlowers(flowerbed, n):
    count = 0
    length = len(flowerbed)
    
    for i in range(length):
        if flowerbed[i] == 0:
            # Check if the previous and next spots are empty or out of bounds
            prev_empty = (i == 0 or flowerbed[i - 1] == 0)
            next_empty = (i == length - 1 or flowerbed[i + 1] == 0)
            
            if prev_empty and next_empty:
                flowerbed[i] = 1  # Plant a flower
                count += 1
        
        if count >= n:
            return True
            
    return count >= n

Example Usage

To use the canPlaceFlowers function, we need to set up a flowerbed and the number of flowers we want to plant:

flowerbed = [1, 0, 0, 0, 1]
n = 1
result = canPlaceFlowers(flowerbed, n)
print(result)  # Output: True

This code checks the flowerbed for places to put flowers. It follows the rules well. For more similar problems, we can look at articles like Array Two Sum or Array Best Time to Buy and Sell Stock.

Optimized Greedy Solution in Python for Can Place Flowers

The Optimized Greedy Solution for the “Can Place Flowers” problem is about going through the flowerbed array. We check for the right spots to plant flowers. We make sure that no two flowers are next to each other. This method works in a linear time of O(n).

Algorithm Steps:

1. Start with a count of flowers we can plant.
2. Use a loop to go through the flowerbed.
3. Check if each spot is empty (0) and if the spots next to it are also empty or out of bounds.
4. If the spot is good, plant a flower by setting it to 1 and add to the count.
5. Return the total count of flowers we planted.

Python Implementation:

def canPlaceFlowers(flowerbed, n):
    count = 0
    length = len(flowerbed)
    
    for i in range(length):
        # Check if the current spot is empty and adjacent spots are empty or out of bounds
        if flowerbed[i] == 0 and (i == 0 or flowerbed[i - 1] == 0) and (i == length - 1 or flowerbed[i + 1] == 0):
            flowerbed[i] = 1  # Plant a flower
            count += 1
        
        if count >= n:  # Early exit if enough flowers are planted
            return True
            
    return count >= n  # Final check for the number of flowers planted

# Example usage
flowerbed = [1, 0, 0, 0, 1]
n = 1
print(canPlaceFlowers(flowerbed, n))  # Output: True

This way, we make sure to plant flowers only when there is enough space. We follow the problem’s rules. The greedy method checks each position in the flowerbed and counts the flowers we plant.

For more problems with arrays, we can look at articles like Array Two Sum or Array Remove Duplicates.

C++ Brute Force Method for Can Place Flowers

The brute force way for the “Can Place Flowers” problem is to check every spot in the flowerbed. We want to see if we can plant a flower without breaking the rules. The rules say that we can’t plant two flowers next to each other.

Implementation

In our C++ code, we go through the flowerbed array. We check each spot. If a spot is empty (0) and the spots next to it are either empty or outside the flowerbed, we can plant a flower (1). The function will return true if we can plant the needed number of flowers. If not, it will return false.

#include <vector>
using namespace std;

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int count = 0;
        int size = flowerbed.size();

        for (int i = 0; i < size; i++) {
            // Check if the current plot is empty and the adjacent plots are empty or out of bounds
            if (flowerbed[i] == 0 &&
                (i == 0 || flowerbed[i - 1] == 0) &&
                (i == size - 1 || flowerbed[i + 1] == 0)) {
                flowerbed[i] = 1; // Plant a flower
                count++;
            }
        }
        return count >= n; // Check if we can plant at least n flowers
    }
};

Explanation of the Code

• Input: The function takes a vector called flowerbed and an integer n. This integer is how many flowers we want to plant.
• Logic:
  • We loop through each spot in the flowerbed.
  • We check if the current spot is empty and if the spots next to it can allow planting.
  • If we plant a flower, we increase the count.
• Output: The function returns true if the count of planted flowers is equal to or more than n.

This brute force method has time complexity of O(m). Here, m is the length of the flowerbed array. We check each spot one time. This method is simple but can be slow for big flowerbeds and many flowers to plant. If we want a faster solution, we can think about using a greedy algorithm.

For more reading on working with arrays, we can check Array Contains Duplicate and Array Maximum Subarray.

C++ Optimized Greedy Algorithm for Can Place Flowers

The “Can Place Flowers” problem asks us to find the most flowers we can plant in a flowerbed. We cannot plant flowers next to each other. The optimized greedy algorithm gives us a fast way to solve this problem.

Algorithm Explanation

1. Input: We have an array that shows the flowerbed. In this array, 0 means an empty spot and 1 means a flower is already planted. We also have an integer n that tells us how many flowers we want to plant.
2. Greedy Strategy: We loop through the flowerbed. At each position, we check if we can plant a flower. We can plant a flower at position i if:
   • The current position is empty (flowerbed[i] == 0)
   • The spots next to it are also empty or out of bounds (flowerbed[i-1] == 0 and flowerbed[i+1] == 0).

C++ Code Implementation

#include <vector>
using namespace std;

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int count = 0;
        int size = flowerbed.size();
        
        for (int i = 0; i < size; i++) {
            // Check if the current spot is empty
            if (flowerbed[i] == 0) {
                // Check adjacent spots
                bool isLeftEmpty = (i == 0 || flowerbed[i - 1] == 0);
                bool isRightEmpty = (i == size - 1 || flowerbed[i + 1] == 0);
                
                // If both adjacent spots are empty, we can plant a flower
                if (isLeftEmpty && isRightEmpty) {
                    flowerbed[i] = 1; // Plant the flower
                    count++; // Increase the count
                }
            }
            // Exit early if we have planted enough flowers
            if (count >= n) {
                return true;
            }
        }
        return count >= n;
    }
};

Complexity Analysis

• Time Complexity: O(m). Here, m is the size of the flowerbed array. We go through the array one time.
• Space Complexity: O(1). We use a small amount of extra space.

This optimized greedy algorithm helps us find out if we can plant n flowers in the flowerbed. If you want to learn more about problems with arrays, you can look at Array Contains Duplicate or Array Maximum Subarray.

Time Complexity Analysis of Array Can Place Flowers Solutions

In the “Can Place Flowers” problem, we want to find out how many flowers we can plant in a flowerbed shown as an array. We must make sure that no two flowers are next to each other. The time complexity of different solutions changes depending on the method we use.

Brute Force Approach

The brute force approach looks at every possible spot to plant a flower. It goes through the whole flowerbed array. The time complexity for this method is:

• Time Complexity: O(n^2)

Here, n means the length of the flowerbed array. For each spot, we might need to check the spots next to it. This creates a nested loop.

Optimized Greedy Algorithm

The optimized greedy approach goes through the flowerbed array. It checks if we can plant a flower at the current spot based on some rules. The spot must be empty, and the spots next to it must also be empty. This method only needs to go through the array once.

• Time Complexity: O(n)

This linear time complexity comes from going through the flowerbed array just one time. This makes this method much better than the brute force method.

Summary of Time Complexities

• Brute Force Approach: O(n^2)
• Optimized Greedy Algorithm: O(n)

In summary, we prefer the optimized greedy algorithm because it has linear time complexity. This makes it good for bigger inputs in the “Can Place Flowers” problem. This shows how important it is to choose the right algorithm based on the problem we have.

For more reading about array problems, we can check out articles like Array Two Sum - Easy or Array Contains Duplicate - Easy.

Frequently Asked Questions

1. What is the goal of the “Can Place Flowers” problem?

The “Can Place Flowers” problem is a common challenge in algorithms. The goal is to see if we can plant a certain number of flowers in a flowerbed. We must not plant two flowers next to each other. We can solve this problem using different methods. These methods include brute force and greedy algorithms. We will talk about these methods in this article.

2. How can I solve the “Can Place Flowers” problem using a brute force approach?

To solve the “Can Place Flowers” problem with a brute force approach, we need to go through the flowerbed array. We check every spot to plant a flower. For each spot, we must make sure the spots next to it are empty. This method is simple but can be slow for big arrays. For a clear code example in Java, look at the brute force part of this article.

3. What is a greedy algorithm, and how does it apply to the “Can Place Flowers” problem?

A greedy algorithm is a way to build a solution step by step. We always pick the next step that gives us the most benefit right away. In the “Can Place Flowers” problem, a greedy algorithm helps us find the most flowers we can plant without breaking the no-adjacency rule. We check the empty spaces in one go through the flowerbed. For a full code example in Python, see our greedy solution section.

4. What are the time complexities of the brute force and greedy solutions for “Can Place Flowers”?

The time complexity for the brute force solution of the “Can Place Flowers” problem is O(n^2). This is because we may need to check each spot against every other spot in the worst case. On the other hand, the greedy solution works in O(n) time. This makes it much faster, especially for big inputs. This speed is important when we solve similar problems with arrays.

5. Can I find more array-related algorithm problems similar to “Can Place Flowers”?

Yes! There are many problems like “Can Place Flowers” in array algorithms. For example, you can look at “Array: Best Time to Buy and Sell Stock” or “Array: Contains Duplicate” problems. These problems help us learn more about arrays and how algorithms work. Check these articles for more details: Best Time to Buy and Sell Stock and Contains Duplicate.