Problem Description

The "Rabbits in Forest" problem asks you to determine the minimum number of rabbits that could be present in a forest based on the answers you get from some rabbits. Each rabbit is asked: "How many other rabbits have the same color as you?" The answers are collected in an integer array answers, where each element represents one rabbit's reply.

Each answer answers[i] means that this rabbit claims there are answers[i] other rabbits with the same color, so at least answers[i] + 1 rabbits of that color exist. Multiple rabbits can give the same answer, but their responses might refer to the same group or different groups.

Your task is to find the minimum total number of rabbits that could be present in the forest, given these answers. You cannot assume that all rabbits with the same answer belong to the same group, as there could be more than one group of the same size.

• Each answers[i] is an integer in the range 0 to 1000.
• Do not reuse rabbits between groups unless logically necessary.
• There is always at least one valid solution.

Thought Process

At first glance, it might seem that if several rabbits give the same answer, they must all be describing a single group. However, that's not always true! If more rabbits give the same answer than the group size allows, they must be split into multiple groups.

For example, if three rabbits say "2" (meaning "there are 2 others like me"), that means there are at least three rabbits of that color. But if five rabbits say "2", they can't all be in the same group, because a group of size 3 can only contain three rabbits. So, the extra rabbits must form another group.

To optimize, we need to group rabbits with the same answer together, and for each group, calculate how many groups are needed to cover all the rabbits giving that answer, without exceeding the group size.

Brute-force would try all possible groupings, but that's inefficient. Instead, we can use a hashmap to count answers and do the math for the minimal number of groups required for each answer.

Solution Approach

We can break the solution into these steps:

1. Count the answers: Use a hash map (or dictionary) to count how many times each answer appears in the answers array.
2. Process each unique answer: For each unique answer k, we know each group of rabbits with that answer must have size k + 1 (since each rabbit says there are k others).
3. Calculate number of groups: If there are c rabbits giving answer k, the minimum number of such groups is ceil(c / (k + 1)). This ensures we don't put more than k + 1 rabbits in a group.
4. Sum up the rabbits: For each group, add number of groups * (k + 1) to the total, as that's the minimum number of rabbits needed for those answers.

The use of a hash map allows us to efficiently count and process each answer in linear time.

Example Walkthrough

Let's walk through an example with answers = [1, 1, 2].

• Count of answers: 1 appears twice, 2 appears once.
• For answer 1 (meaning each rabbit says "1 other rabbit has my color"): group size is 1 + 1 = 2. We have 2 rabbits, which fits exactly into 1 group of size 2, so we need 2 rabbits for this group.
• For answer 2: group size is 2 + 1 = 3. Only 1 rabbit says this, but a group must have 3 rabbits. So, we need to assume there are 3 such rabbits (the other 2 may not have answered).
• Total rabbits = 2 (from answer 1) + 3 (from answer 2) = 5.

This process ensures we don't underestimate the number of rabbits and always use the minimal possible total.

Code Implementation

from collections import Counter
import math

class Solution:
    def numRabbits(self, answers):
        count = Counter(answers)
        res = 0
        for k, c in count.items():
            group_size = k + 1
            num_groups = math.ceil(c / group_size)
            res += num_groups * group_size
        return res
      

#include <vector>
#include <unordered_map>
#include <cmath>
using namespace std;

class Solution {
public:
    int numRabbits(vector<int>& answers) {
        unordered_map<int, int> count;
        for (int a : answers) count[a]++;
        int res = 0;
        for (auto& [k, c] : count) {
            int group_size = k + 1;
            int num_groups = (c + group_size - 1) / group_size;
            res += num_groups * group_size;
        }
        return res;
    }
};
      

import java.util.*;

class Solution {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> count = new HashMap<>();
        for (int a : answers) count.put(a, count.getOrDefault(a, 0) + 1);
        int res = 0;
        for (Map.Entry<Integer, Integer> entry : count.entrySet()) {
            int k = entry.getKey();
            int c = entry.getValue();
            int groupSize = k + 1;
            int numGroups = (c + groupSize - 1) / groupSize;
            res += numGroups * groupSize;
        }
        return res;
    }
}
      

var numRabbits = function(answers) {
    const count = new Map();
    for (let a of answers) {
        count.set(a, (count.get(a) || 0) + 1);
    }
    let res = 0;
    for (let [k, c] of count.entries()) {
        let groupSize = k + 1;
        let numGroups = Math.ceil(c / groupSize);
        res += numGroups * groupSize;
    }
    return res;
};
      

Time and Space Complexity

Brute-force: A brute-force approach would try all possible groupings of rabbits, which is exponential in the number of answers and not feasible for large inputs.

Optimized Solution:

• Time Complexity: O(n), where n is the number of elements in answers. We traverse the array to count occurrences and then process each unique answer.
• Space Complexity: O(m), where m is the number of unique answers (at most 1001 due to constraints). We store counts in a hash map.

Summary

The "Rabbits in Forest" problem is a clever application of grouping and counting logic, where the key insight is to recognize that rabbits giving the same answer may belong to different groups. By using a hash map to count answers and carefully calculating the minimum number of groups for each answer, we arrive at an elegant and efficient solution. This approach avoids brute-force grouping and leverages mathematical reasoning for minimal total rabbits.