What is the time complexity of Rabbits in Forest?

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

What data structures are used to solve Rabbits in Forest?

This problem uses Array, Hash Table, Math, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Rabbits in Forest?

Insight 1: Rabbits answering the same number could belong to the same color group. Insight 2: If more rabbits answer a number 'x' than the group size (x+1), then we need to form multiple groups of size (x+1).

Which companies ask LeetCode 781. Rabbits in Forest in interviews?

This problem is frequently asked at CARS24, Cleartrip, Wish, Zepto.

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Rabbits in Forest - LeetCode 781 Solution

Rabbits in Forest - Complete Solution Guide

Rabbits in Forest is LeetCode problem 781, 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

There is a forest with an unknown number of rabbits. We asked n rabbits "How many rabbits have the same color as you?" and collected the answers in an integer array answers where answers[i] is the answer of the i th rabbit. Given the array answers , return the minimum number of rabbits that could be in the forest . Example 1: Input: answers = [1,1,2] Output: 5 Explanation: The two rabbits that answered "1" could both be the same color, say red. The rabbit that answered "2" can't be red or the an

Detailed Explanation

The problem asks us to determine the minimum number of rabbits in a forest based on the answers given by some of the rabbits. Each rabbit provides a number 'x', indicating that there are 'x' other rabbits with the same color as itself. The goal is to find the smallest possible total number of rabbits in the forest, considering all the answers.

Solution Approach

The solution utilizes a counting approach. First, it counts the occurrences of each answer. Then, for each distinct answer, it calculates the number of groups needed and, thus, the total number of rabbits belonging to those groups. The sum of rabbits across all distinct answers gives the minimum possible total number of rabbits in the forest.

Step-by-Step Algorithm

Step 1: Count the occurrences of each answer in the `answers` array using a hash map (or `collections.Counter` in Python).
Step 2: Iterate through the distinct answers and their counts in the hash map.
Step 3: For each distinct answer `k` (representing the number of other rabbits of the same color) and its count `c`, calculate the group size as `group_size = k + 1`.
Step 4: Determine the number of groups required by calculating `num_groups = (c + group_size - 1) // group_size`. This uses ceiling division to round up to the nearest whole number.
Step 5: Add `num_groups * group_size` to the `total_rabbits` count.
Step 6: After processing all distinct answers, return the `total_rabbits`.

Key Insights

Insight 1: Rabbits answering the same number could belong to the same color group.
Insight 2: If more rabbits answer a number 'x' than the group size (x+1), then we need to form multiple groups of size (x+1).
Insight 3: The number of rabbits in the forest is minimized when we group as many rabbits with the same answer together as possible.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Math, Greedy.

Companies

Asked at: CARS24, Cleartrip, Wish, Zepto.

Rabbits in Forest | LeetCode 781

Medium

Array Hash Table Math Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> counts = new HashMap<>();
        for (int answer : answers) {
            counts.put(answer, counts.getOrDefault(answer, 0) + 1);
        }

        int total_rabbits = 0;
        for (Map.Entry<Integer, Integer> entry : counts.entrySet()) {
            int k = entry.getKey();
            int c = entry.getValue();
            int group_size = k + 1;
            int num_groups = (c + group_size - 1) / group_size;
            total_rabbits += num_groups * group_size;
        }
        return total_rabbits;
    }
}

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Problem Info

Description

Problem 781: Rabbits in Forest - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Greedy

Companies

CARS24CleartripWishZepto

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:781

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

6 steps

1
Step 1: Count the occurrences of each answer in the `answers` array using a hash map (or `collections.Counter` in Python).
2
Step 2: Iterate through the distinct answers and their counts in the hash map.
3
Step 3: For each distinct answer `k` (representing the number of other rabbits of the same color) and its count `c`, calculate the group size as `group_size = k + 1`.
4
Step 4: Determine the number of groups required by calculating `num_groups = (c + group_size - 1) // group_size`. This uses ceiling division to round up to the nearest whole number.
5
Step 5: Add `num_groups * group_size` to the `total_rabbits` count.
6
Step 6: After processing all distinct answers, return the `total_rabbits`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the iterations through the input array and the hash map. The space complexity is determined by the hash map, which stores the frequency of each answer. Both are linearly dependent on the input size.

📖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

Sorting

Sort the array and iterate through it. If `answers[i]` is equal to `answers[i+1]`, consider them in the same group. If not, increment the rabbit count by `answers[i] + 1`. This approach is conceptually similar but sorting adds overhead.

Time: O(n log n) due to sorting.Space: O(1) if sorting in place, O(n) if sorting requires auxiliary space.

Trade-offs: Sorting is slower than using a hash map due to the O(n log n) time complexity. It might be preferable if space complexity is a major concern and in-place sorting is possible, but the hash map approach is generally faster.

✨

Key Insights

1Insight 1: Rabbits answering the same number could belong to the same color group.
2Insight 2: If more rabbits answer a number 'x' than the group size (x+1), then we need to form multiple groups of size (x+1).
3Insight 3: The number of rabbits in the forest is minimized when we group as many rabbits with the same answer together as possible.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Empty input array. The algorithm correctly returns 0 in this case, as there are no rabbits.
Edge case 2: All rabbits give the same answer. The algorithm correctly calculates the number of groups and total rabbits required in this scenario.
Edge case 3: All rabbits give distinct answers. Each answer represents a group of rabbits of a unique color, and the total number of rabbits is calculated accordingly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not accounting for multiple groups with the same answer when the count of an answer exceeds the group size (k+1).

✗Common mistake 2: Assuming that all rabbits giving the same answer are necessarily the same color. The logic should only group as many rabbits as the size of group allows, and separate any excess.

✗Common mistake 3: Integer division issues, ensuring the 'num_groups' calculation does not truncate values which require another full group.

✅

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: Estimating population sizes based on survey responses where individuals indicate the number of other individuals they know within a specific category.

🏦Real-world scenario 2: Network analysis, where nodes report the number of neighbors they have, and the goal is to estimate the minimum total number of nodes in the network.

💡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

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 1, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Rabbits in Forest"?

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

🤖 Need help with this problem?