Rabbit Math: When Saying “1” Doesn’t Mean Just One

Minimum Rabbits in the Forest – Decoding the Puzzle

This problem challenges our ability to reason through indirect information. Given a set of answers from rabbits about how many others share their color, our goal is to estimate the minimum number of rabbits that could be present in the forest. Let’s analyze this step-by-step and explore both brute force and optimized solutions with a focus on edge cases and estimation logic.

Problem Statement

You are given an integer array answers where each answers[i] represents a rabbit’s response to the question:
"How many rabbits have the same color as you?"

Your task is to determine the minimum number of rabbits that could be in the forest based on these answers.

Example 1:

  • Input: [1, 1, 2]

  • Output: 5

  • Explanation: The two rabbits answering "1" can be in one group of 2. The rabbit answering "2" implies a group of 3, but only one has been seen, so two more must exist unseen.

Best Data Structure

The most suitable data structure for this problem is a HashMap (or dictionary). It allows efficient frequency counting and helps us group rabbits with the same answer.

Different Approaches

Brute Force Approach

The intuition is to simulate the grouping of rabbits. Each rabbit that says "x" implies a group of x + 1 rabbits of the same color. We create new groups when necessary and avoid reusing group slots beyond capacity.


public int numRabbitsBrute(int[] answers) {
    Map<Integer, Integer> map = new HashMap<>();
    int total = 0;

    for (int ans : answers) {
        if (!map.containsKey(ans) || map.get(ans) == 0) {
            total += (ans + 1);
            map.put(ans, ans); // Reserve group space
        } else {
            map.put(ans, map.get(ans) - 1);
        }
    }
    return total;
}

Optimized Approach

Instead of simulating the process, we use mathematical grouping. For each answer x, the maximum group size is x + 1. If there are multiple rabbits with the same answer, we divide them into groups of size x + 1. The number of groups required is the ceiling of (count / (x + 1)), and the total rabbits for that answer becomes groups * (x + 1).

Java Code – Optimized


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

    int total = 0;
    for (int x : map.keySet()) {
        int count = map.get(x);
        int groupSize = x + 1;
        int groups = (int) Math.ceil((double) count / groupSize);
        total += groups * groupSize;
    }
    return total;
}

⚠️Common Pitfalls

  1. Ignoring the ceiling logic: Without using ceiling division, the count of required groups will be underestimated.

  2. Overusing group space: Reusing a group for more rabbits than allowed leads to invalid results.

  3. Special case for zero: An answer of 0 means the rabbit is alone in its color group. Each 0 counts as one distinct rabbit.

  4. Assuming all same answers belong to one group: This is not always valid; answers must be split into separate groups when their count exceeds the allowed group size

📊Time and Space Complexity Analysis

Both the brute force and optimized solutions run in linear time with respect to the number of elements in the input array. 

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

🌍Real-World Applications

This problem mirrors real-world scenarios where estimates must be made based on partial reports. Applications include:

  • Wildlife population estimation using limited samples.

  • Survey analysis where participants provide partial group information.

  • Modeling distributed systems or social clusters with partial visibility.

I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

Comments

Post a Comment

Popular posts from this blog

Trailing Zeros in Factorial – How Many and Why?

Trailing Zeros – Counting Zeros in Factorial Factorial-related problems are a fundamental part of Data Structures and Algorithms (DSA). One such classic problem involves counting the number of trailing zeros in N! . In this article, we will explore different approaches to solving this problem, optimize our solution, and examine its real-world applications. Optimal Data Structure for Solving the Problem For this problem, no additional data structures are required. Since factorials grow exponentially, storing the actual factorial value is impractical. Instead, we leverage mathematical properties and division operations to efficiently determine the number of trailing zeros. This approach ensures an optimal solution in terms of both time and space complexity. Approach – From Brute Force to an Optimized Solution Brute Force Approach – Compute and Count A naive approach involves computing the factorial of N and counting the number of trailing zeros. However, this method is highly ineff...
Read more

Divisible Drama — Who’s In, Who’s Out?

 Divisible Drama — Who’s In, Who’s Out? Problem Statement You’re given two positive integers n and m . num1 = sum of all numbers from 1 to n not divisible by m num2 = sum of all numbers from 1 to n divisible by m Your mission, should you choose to accept it: Return num1 - num2 Example Input: n = 10 , m = 3 Output: 19 Breakdown: Not divisible by 3 → [ 1 , 2 , 4 , 5 , 7 , 8 , 10 ] → num1 = 37 Divisible by 3 → [ 3 , 6 , 9 ] → num2 = 18 Answer = 37 - 18 = 19 Best Data Structure for Solving the Problem This one’s a math.  Why overthink it when Gauss already gave us the formula for sum of 1 to n? Different Approaches (from Brute Force to Optimized) Approach 1: Brute Force (aka List & Sum Vibes) Loop from 1 to n , split the nums based on whether they’re divisible by m , and sum them up. class Solution { public int differenceOfSums ( int n, int m) { int num1 = 0 , num2 = 0 ; for ( int i = 1 ; i <= n; i++)...
Read more

Zero Array After Queries

Zero Array After Queries You are given an integer array nums and a list of queries, where each query is a pair [li, ri] representing a range. Each query allows you to decrement any subset of elements in the range [li, ri] by 1. Your task is to determine if it is possible to make every element in nums equal to zero after applying all queries in order . Example Input: nums = [ 1 , 0 , 1 ] queries = [[0, 2]] Output: true Explanation: We can select indices 0 and 2 and decrement both by 1 to obtain [0, 0, 0] . Best Data Structure for the Job We are working with multiple range updates and need to evaluate coverage over indices. The best choice here is the difference array combined with a prefix sum . It allows efficient range additions and helps track how many times each index is affected by the queries. Different Approaches Approach 1: Brute Force This naive method iterates over every index for every query and performs the decrement operations directly. class Solution {...
Read more