Problem Summary

You are given an integer array answers, where each answers[i] represents:

A rabbit says:
“There are answers[i] other rabbits that have the same color as me.”

Your task is to compute the minimum number of rabbits that could be in the forest.

Example
Input: answers = [1, 1, 2]
Output: 5
Explanation:

• Two rabbits say “1 other rabbit has my color” → together form a group of 2 rabbits.
• One rabbit says “2 other rabbits have my color” → requires a group size of 3 rabbits.
  Minimum = 2 + 3 = 5.

Example
Input: answers = [10, 10, 10]
Output: 11
Explanation:
Each rabbit says there are 10 rabbits of its color; group size = 11.
We have 3 rabbits saying this, but they can all be in the same group of 11.

Constraints

• 1 ≤ answers.length ≤ 1000
• 0 ≤ answers[i] ≤ 1000

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Key Insight

If a rabbit says “there are x other rabbits of my color”, then the size of its color group must be:

group size = x + 1

For example:

• If a rabbit says 0 → group size = 1
• If a rabbit says 1 → group size = 2
• If a rabbit says 2 → group size = 3
  etc.

If several rabbits give the same number x, they can share the same group, but only up to the allowed group size.

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Approach (Simple English)

1. Count how many rabbits say each value x.
2. For each unique x:
   • The group size is x + 1.
   • If count rabbits gave that answer:
     • They fill groups of size x + 1.
     • Number of groups needed = ceil(count / (x + 1))
   • Total rabbits contributed = number_of_groups * (x + 1)
3. Sum results for all x.

Why This Works

Each answer x forces rabbits into buckets of size x + 1.
If more rabbits claim the same x than one group allows, they must form multiple such groups.

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Java Code (Optimal Solution)

import java.util.*;

class Solution {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> freq = new HashMap<>();

        // Count occurrences
        for (int x : answers) {
            freq.put(x, freq.getOrDefault(x, 0) + 1);
        }

        int total = 0;

        // Process each unique answer
        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
            int x = entry.getKey();
            int count = entry.getValue();

            int groupSize = x + 1;

            // number of groups needed
            int groups = (count + groupSize - 1) / groupSize;

            total += groups * groupSize;
        }

        return total;
    }
}

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Dry Run

Input:

answers = [1, 1, 2]

Step 1: Count frequencies

• 1 → 2 rabbits
• 2 → 1 rabbit

Step 2: Process each x

For x = 1

• group size = 2
• count = 2
• groups = ceil(2 / 2) = 1
• rabbits = 1 * 2 = 2

For x = 2

• group size = 3
• count = 1
• groups = ceil(1 / 3) = 1
• rabbits = 1 * 3 = 3

Total = 2 + 3 = 5

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Complexity Analysis

Time Complexity:
O(n), counting + iteration through map.

Space Complexity:
O(n), for storing frequencies.

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Edge Cases

• All rabbits say 0 → each is a unique color → answer = number of rabbits
• All rabbits say the same x → they fill up groups of size x+1
• Single rabbit → group size = 1
• Large answers (like 1000) work normally since group sizes scale accordingly