What is the time complexity of Can Place Flowers?

The optimal solution has a time complexity of O(m).

What data structures are used to solve Can Place Flowers?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Can Place Flowers?

Insight 1: We can iterate through the flowerbed and check for available spots (represented by '0'). Insight 2: For each empty spot, we need to check if its adjacent plots are also empty.

Which companies ask LeetCode 605. Can Place Flowers in interviews?

This problem is frequently asked at Agoda, Airbnb, Atlassian, Cisco, LinkedIn and 3 more companies.

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Can Place Flowers - LeetCode 605 Solution

Can Place Flowers - Complete Solution Guide

Can Place Flowers is LeetCode problem 605, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You have a long flowerbed in which some of the plots are planted, and some are not. However, flowers cannot be planted in adjacent plots. Given an integer array flowerbed containing 0 's and 1 's, where 0 means empty and 1 means not empty, and an integer n , return true if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule and false otherwise . Example 1: Input: flowerbed = [1,0,0,0,1], n = 1 Output: true Example 2: Input: flowerbed = [1,0,0,0,1], n = 2

Detailed Explanation

The problem asks us to determine if we can plant `n` new flowers in a flowerbed (represented as an array of 0s and 1s) without planting flowers in adjacent plots. A '0' in the array indicates an empty plot, and a '1' indicates a planted plot. The constraint is that no two flowers can be planted next to each other.

Solution Approach

The provided solutions use a greedy approach. They iterate through the flowerbed, and whenever they find an empty plot, they check if the adjacent plots are also empty. If both adjacent plots are empty, they plant a flower in the current plot (by changing the value to 1) and increment the count of planted flowers. Finally, they check if the count of planted flowers is greater than or equal to `n`.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0, representing the number of flowers planted.
Step 2: Iterate through the `flowerbed` array from left to right.
Step 3: For each plot, check if it's empty (flowerbed[i] == 0).
Step 4: If the plot is empty, check if the adjacent plots are also empty. Consider edge cases for the first and last plots.
Step 5: If both adjacent plots are empty (or only one adjacent plot exists and it is empty at the edges), plant a flower in the current plot (flowerbed[i] = 1) and increment the `count`.
Step 6: If `count` is greater than or equal to `n`, return `true` (we've planted enough flowers).
Step 7: After iterating through the entire flowerbed, return `true` if `count` is greater than or equal to `n`; otherwise, return `false`.

Key Insights

Insight 1: We can iterate through the flowerbed and check for available spots (represented by '0').
Insight 2: For each empty spot, we need to check if its adjacent plots are also empty.
Insight 3: Planting a flower at an index updates the flowerbed, potentially affecting future planting decisions, so modifying the array in place is fine, but requires careful consideration.
Insight 4: The edges of the flowerbed require special handling since there's only one adjacent plot to check.

Complexity Analysis

Time Complexity: O(m)

Space Complexity: O(m)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Agoda, Airbnb, Atlassian, Cisco, LinkedIn, Nike, SOTI, Yandex.

Can Place Flowers | LeetCode 605

Easy

Array Greedy

Time: O(m)Space: O(m)

Solution Code

Available in:

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int count = 0;
        int i = 0;
        while (i < flowerbed.length) {
            if (flowerbed[i] == 0) {
                boolean canPlace = true;
                if (i > 0 && flowerbed[i - 1] == 1) {
                    canPlace = false;
                }
                if (i < flowerbed.length - 1 && flowerbed[i + 1] == 1) {
                    canPlace = false;
                }
                if (canPlace) {
                    flowerbed[i] = 1;
                    count++;
                }
            }
            if (count >= n) {
                return true;
            }
            i++;
        }
        return count >= n;
    }
}

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Problem Info

Description

Problem 605: Can Place Flowers - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

AgodaAirbnbAtlassianCiscoLinkedInNikeSOTIYandex

Stats

Difficulty:Easy

Languages:4

Companies:8

Problem #:605

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

7 steps

1
Step 1: Initialize a counter `count` to 0, representing the number of flowers planted.
2
Step 2: Iterate through the `flowerbed` array from left to right.
3
Step 3: For each plot, check if it's empty (flowerbed[i] == 0).
4
Step 4: If the plot is empty, check if the adjacent plots are also empty. Consider edge cases for the first and last plots.
5
Step 5: If both adjacent plots are empty (or only one adjacent plot exists and it is empty at the edges), plant a flower in the current plot (flowerbed[i] = 1) and increment the `count`.
6
Step 6: If `count` is greater than or equal to `n`, return `true` (we've planted enough flowers).
7
Step 7: After iterating through the entire flowerbed, return `true` if `count` is greater than or equal to `n`; otherwise, return `false`.

💡

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(m)

💾

Space Complexity

O(m)

📝

Why?

The time complexity is directly proportional to the size of the input array because the algorithm needs to visit each element at most once. The space complexity is constant since we're only using a few extra variables (like `count` and `i`) regardless of the input array's size. In the python solution, the space complexity is O(m) because it adds extra elements to the flowerbed in the beginning.

📖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

No alternative name, but logic change only

Instead of modifying the array, create a copy and modify the copy. The original array is untouched. This prevents side-effects if the original array is required elsewhere.

Time: O(m)Space: O(m)

Trade-offs: Uses extra memory, but ensures the original array is unchanged.

✨

Key Insights

1Insight 1: We can iterate through the flowerbed and check for available spots (represented by '0').
2Insight 2: For each empty spot, we need to check if its adjacent plots are also empty.
3Insight 3: Planting a flower at an index updates the flowerbed, potentially affecting future planting decisions, so modifying the array in place is fine, but requires careful consideration.
4Insight 4: The edges of the flowerbed require special handling since there's only one adjacent plot to check.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Empty flowerbed (flowerbed.length == 0). In this case, if n is 0 return true. If n > 0 return false (all implementations handle this correctly, either implicitly or explicitly).
Edge case 2: Flowerbed with only one plot (flowerbed.length == 1). If the plot is empty (0) and n is 1, we can plant a flower. If the plot is already planted (1), we cannot plant any new flowers.
Edge case 3: 'n' is zero (n == 0). In this case, we can always plant 0 flowers, so the result should always be true (all implementations handle this correctly).
Edge case 4: Flowerbed starts and ends with zero. The greedy approach takes advantage of this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases for the first and last plots correctly. It's important to treat the ends of the flowerbed as if there's an empty plot beyond them.

✗Common mistake 2: Not updating the flowerbed after planting a flower. If you don't update flowerbed[i] to 1 after planting, the next iteration might incorrectly assume it's still empty.

✅

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 in a grid. Imagine allocating servers in a data center rack, where servers cannot be placed next to each other due to cooling requirements. This problem helps determine if we can allocate a given number of servers.

🏦Real-world scenario 2: Planning seating arrangements in a theater or airplane, where certain seats must be kept empty between occupied seats for accessibility or comfort reasons. This algorithm can help determine if a certain number of people can be seated.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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 5, 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

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

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Can Place Flowers"?

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

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