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Leetcode 781. Rabbits in Forest

Problem Statement:

In a forest, each rabbit has some color. Some rabbits tell you how many other rabbits have the same color as them. You are given an integer array answers, where answers[i] is the answer of the i-th rabbit.

Return the minimum number of rabbits that could be in the forest.

Clarifying Questions:

  1. What does answers[i] == 0 mean?
    • If answers[i] == 0, it means that the rabbit claims it is the only rabbit with that color.
  2. Are there any constraints or limits on the length of the answers array or the numbers within it?
    • The array answers will have a length in the range [1, 1000], and each element will be an integer in the range [0, 1000].
  3. Are we provided any additional context about the colors or should we infer it purely based on the answers?
    • We should infer the number of rabbits purely based on the answers array without additional context about the colors.

Strategy:

  1. Group the rabbits by their answers: We’ll use a hashmap to count how many rabbits gave each particular answer.
  2. Calculate the number of rabbits for each group:
    • For each distinct answer y, if there are x rabbits that say there are y other rabbits of the same color, this means there are at most y + 1 rabbits of that color.
    • If there are more than y + 1 such rabbits, those must overflow into another group of y + 1 rabbits.
    • We can calculate the minimum number of rabbits needed for each group by using the ceiling of the division of the rabbit count by y + 1 (i.e., (count + y) // (y + 1) * (y + 1)).

Code:

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

public class RabbitsInForest {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> answerCount = new HashMap<>();
        
        // Count the occurrences of each answer
        for (int answer : answers) {
            answerCount.put(answer, answerCount.getOrDefault(answer, 0) + 1);
        }

        int minRabbits = 0;

        // Calculate the minimum number of rabbits for each answer group
        for (Map.Entry<Integer, Integer> entry : answerCount.entrySet()) {
            int y = entry.getKey();
            int count = entry.getValue();
            
            // Calculate group size which is y + 1
            int groupSize = y + 1;
            
            // Calculate the number of groups needed
            int groups = (count + groupSize - 1) / groupSize;
            
            // Add the total number of rabbits in these groups
            minRabbits += groups * groupSize;
        }
        
        return minRabbits;
    }

    public static void main(String[] args) {
        RabbitsInForest rif = new RabbitsInForest();
        int[] answers = {1, 1, 2};
        System.out.println(rif.numRabbits(answers));  // Output: 5
    }
}

Time Complexity:

This efficient approach ensures that we capture the least number of rabbits as inferred from their answers while adhering to the logic provided.

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