Approach and Intuition

The problem revolves around determining the minimum number of rabbits given the number of rabbits of the same color each rabbit claims to have seen. The key observation is that if a rabbit answers x, then there are x + 1 rabbits of that color (including the rabbit itself).

We can use a hash map (or dictionary in Python) to count the occurrences of each answer. For each answer x, we have count rabbits that gave that answer. We need to find the minimum number of groups of x + 1 rabbits to accommodate these count rabbits. This can be calculated using ceiling division (ceil(count / (x + 1))). The total number of rabbits for this answer x is then ceil(count / (x + 1)) * (x + 1). We sum this up for all answers to get the minimum number of rabbits.

Algorithm

1. Count Answers: Create a hash map (or dictionary) to store the frequency of each answer.
2. Calculate Minimum Rabbits per Answer: For each answer x and its frequency count:
   • Calculate the number of groups needed: ceil(count / (x + 1))
   • Calculate the total number of rabbits for this answer: ceil(count / (x + 1)) * (x + 1)
3. Sum up the Rabbits: Sum up the total rabbits calculated for all answers to get the minimum total number of rabbits.

Code Implementation

Python

import math
from collections import Counter
 
class Solution:
    def numRabbits(self, answers: List[int]) -> int:
        counter = Counter(answers)
        total_rabbits = 0
        for answer, count in counter.items():
            total_rabbits += math.ceil(count / (answer + 1)) * (answer + 1)
        return total_rabbits
 

Java

import java.util.HashMap;
import java.util.Map;
 
class Solution {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int answer : answers) {
            counter.put(answer, counter.getOrDefault(answer, 0) + 1);
        }
        int totalRabbits = 0;
        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
            int answer = entry.getKey();
            int count = entry.getValue();
            totalRabbits += (int) Math.ceil((double) count / (answer + 1)) * (answer + 1);
        }
        return totalRabbits;
    }
}

Time and Space Complexity Analysis

Time Complexity: O(N), where N is the length of the answers array. This is because we iterate through the array once to count the answers and then iterate through the hash map (which has at most N entries), both taking linear time.

Space Complexity: O(K), where K is the number of unique answers. In the worst case, K can be N, but it is usually much smaller. The space is used to store the hash map.