What is the time complexity of Flower Planting With No Adjacent?

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

What is the space complexity of Flower Planting With No Adjacent?

The space complexity is O(n + m).

What data structures are used to solve Flower Planting With No Adjacent?

This problem uses Depth-First Search, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Flower Planting With No Adjacent?

Insight 1: The problem can be modeled as a graph coloring problem where gardens are nodes and paths are edges. We need to color the graph with 4 colors such that no two adjacent nodes have the same color. Insight 2: Since each garden has at most 3 neighbors, we can always find a valid flower type for each garden by checking the colors of its neighbors. Because there are 4 flower types, we are guaranteed to find one not used by neighbors.

Which companies ask LeetCode 1042. Flower Planting With No Adjacent in interviews?

This problem is frequently asked at Vimeo.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Flower Planting With No Adjacent - LeetCode 1042 Solution

Flower Planting With No Adjacent - Complete Solution Guide

Flower Planting With No Adjacent is LeetCode problem 1042, a Medium 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 n gardens, labeled from 1 to n , and an array paths where paths[i] = [x i , y i ] describes a bidirectional path between garden x i to garden y i . In each garden, you want to plant one of 4 types of flowers. All gardens have at most 3 paths coming into or leaving it. Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers. Return any such a choice as an array answer , where answer[i] is the type of fl

Detailed Explanation

The problem asks us to assign one of four flower types to each of 'n' gardens, such that no two adjacent gardens (connected by a path) have the same flower type. We are given the number of gardens 'n' and a list of paths 'paths' representing the connections between the gardens. We are guaranteed that a solution exists and each garden has at most 3 neighbors. The output should be an array of length 'n', where the i-th element represents the flower type planted in the (i+1)-th garden.

Solution Approach

The provided solution uses a straightforward greedy approach. It first builds an adjacency list (graph) representation of the garden connections. Then, for each garden, it iterates through the available flower types (1 to 4) and selects the first flower type that is not used by any of its neighbors. This ensures that no two adjacent gardens have the same flower type.

Step-by-Step Algorithm

Step 1: Create an adjacency list (graph) to represent the connections between gardens. The graph[i] will store a list of all gardens connected to garden 'i+1'.
Step 2: Initialize an array 'answer' of size 'n' to store the flower type assigned to each garden. Initially, all elements are set to 0.
Step 3: Iterate through each garden from 0 to n-1.
Step 4: For each garden 'i', create a set (or array in C) 'neighbor_colors' to store the flower types used by its neighbors.
Step 5: Iterate through the neighbors of garden 'i' and add their flower types to the 'neighbor_colors' set if they have already been assigned a flower type (answer[neighbor] != 0).
Step 6: Iterate through the flower types from 1 to 4.
Step 7: If a flower type is not present in the 'neighbor_colors' set, assign that flower type to garden 'i' in the 'answer' array (answer[i] = flower_type) and break the inner loop.
Step 8: After assigning a flower type to all gardens, return the 'answer' array.

Key Insights

Insight 1: The problem can be modeled as a graph coloring problem where gardens are nodes and paths are edges. We need to color the graph with 4 colors such that no two adjacent nodes have the same color.
Insight 2: Since each garden has at most 3 neighbors, we can always find a valid flower type for each garden by checking the colors of its neighbors. Because there are 4 flower types, we are guaranteed to find one not used by neighbors.
Insight 3: The order in which we assign flower types to the gardens doesn't matter as long as we ensure that no adjacent gardens have the same flower type.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Graph.

Companies

Asked at: Vimeo.

Flower Planting With No Adjacent | LeetCode 1042

Medium

Depth-First Search Breadth-First Search Graph

Time: O(n + m)Space: O(n + m)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int[] gardenNoAdj(int n, int[][] paths) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] path : paths) {
            int x = path[0];
            int y = path[1];
            graph.get(x - 1).add(y - 1);
            graph.get(y - 1).add(x - 1);
        }

        int[] answer = new int[n];

        for (int i = 0; i < n; i++) {
            Set<Integer> neighborColors = new HashSet<>();
            for (int neighbor : graph.get(i)) {
                if (answer[neighbor] != 0) {
                    neighborColors.add(answer[neighbor]);
                }
            }

            for (int flowerType = 1; flowerType <= 4; flowerType++) {
                if (!neighborColors.contains(flowerType)) {
                    answer[i] = flowerType;
                    break;
                }
            }
        }

        return answer;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1042: Flower Planting With No Adjacent - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph

Companies

Vimeo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1042

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

8 steps

1
Step 1: Create an adjacency list (graph) to represent the connections between gardens. The graph[i] will store a list of all gardens connected to garden 'i+1'.
2
Step 2: Initialize an array 'answer' of size 'n' to store the flower type assigned to each garden. Initially, all elements are set to 0.
3
Step 3: Iterate through each garden from 0 to n-1.
4
Step 4: For each garden 'i', create a set (or array in C) 'neighbor_colors' to store the flower types used by its neighbors.
5
Step 5: Iterate through the neighbors of garden 'i' and add their flower types to the 'neighbor_colors' set if they have already been assigned a flower type (answer[neighbor] != 0).
6
Step 6: Iterate through the flower types from 1 to 4.
7
Step 7: If a flower type is not present in the 'neighbor_colors' set, assign that flower type to garden 'i' in the 'answer' array (answer[i] = flower_type) and break the inner loop.
8
Step 8: After assigning a flower type to all gardens, return the 'answer' array.

💡

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 + m)

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is determined by the graph construction and the flower assignment process. The space complexity is due to the storage required for the graph and the result array. Since each node has at most 3 edges, m can be considered to be bound by 3n and can be expressed as O(n).

📖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

Depth-First Search (DFS)

You could potentially use DFS to traverse the graph and assign flower types. However, due to the constraint that each garden has at most 3 neighbors, a greedy approach is simpler and more efficient.

Time: O(n + m)Space: O(n + m)

Trade-offs: DFS might be more complex to implement than the greedy approach for this specific problem, without offering any significant performance benefits.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph coloring problem where gardens are nodes and paths are edges. We need to color the graph with 4 colors such that no two adjacent nodes have the same color.
2Insight 2: Since each garden has at most 3 neighbors, we can always find a valid flower type for each garden by checking the colors of its neighbors. Because there are 4 flower types, we are guaranteed to find one not used by neighbors.
3Insight 3: The order in which we assign flower types to the gardens doesn't matter as long as we ensure that no adjacent gardens have the same flower type.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: n = 1, paths = []. This case represents a single garden with no connections. The algorithm correctly assigns flower type 1 to this garden.
Edge case 2: Gardens with no paths. Gardens not part of any path are automatically assigned flower type 1 because their neighbor list is empty.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not creating the adjacency list correctly, which can lead to incorrect neighbor information and invalid flower type assignments.

✗Common mistake 2: Not handling the case where a garden has no neighbors. In this case, any flower type can be assigned to the garden.

✅

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 where resources need to be assigned to entities such that adjacent entities do not share the same resource. For instance, assigning radio frequencies to transmitters to avoid interference.

🏦Real-world scenario 2: Scheduling problems where tasks need to be scheduled in such a way that dependent tasks are not scheduled at the same time. This can be seen as coloring a graph where tasks are nodes and dependencies are edges.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: October 21, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Flower Planting With No Adjacent"?

For this medium-level Depth-First Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?