Problem Informations

• Problem Index: 781
• Problem Link: Rabbits in Forest
• Topics: Array, Hash Table, Math, Greedy

Problem Description

There is a forest with an unknown number of rabbits. We asked n rabbits “How many rabbits have the same color as you?” and collected the answers in an integer array answers where answers[i] is the answer of the $i^{th}$ rabbit.

Given the array answers, return the minimum number of rabbits that could be in the forest.

Example 1:

Input: answers = [1,1,2]
Output: 5
Explanation:
The two rabbits that answered "1" could both be the same color, say red.
The rabbit that answered "2" can't be red or the answers would be inconsistent.
Say the rabbit that answered "2" was blue.
Then there should be 2 other blue rabbits in the forest that didn't answer into the array.
The smallest possible number of rabbits in the forest is therefore 5: 3 that answered plus 2 that didn't.

Example 2:

Input: answers = [10,10,10]
Output: 11

Constraints:

• $1 \leq \text{{answers.length}} \leq 1000$
• $0 \leq \text{{answers[i]}} < 1000$

Intuition

The problem can be viewed as a grouping issue where each unique answer represents a group of rabbits of the same color. The crux is to understand that rabbits answering the same number could potentially belong to the same color group. Every rabbit that provides the same answer belongs to a group that contains itself and other rabbits of the same color which did not answer. The task is to calculate the minimum number of rabbits based on these responses by accounting not only for those that answered but also for those that must exist according to the implications of each answer.

Approach

To solve this problem, we follow these steps:

1. Use a Map to Count Frequencies:

   • Traverse through the given answers array, and for each answer, update a map (countMap) to count occurrences of each distinct answer. This allows us to know how many rabbits provided the same response.

2. Calculate Minimum Rabbits:

   • Iterate over each entry in countMap. Each entry consists of answer (the number as stated by the rabbit) and count (number of rabbits giving that answer).
   • For each distinct answer, we know that all rabbits providing that answer indicate they are part of a group consisting of answer + 1 rabbits (itself plus others of the same color that did not answer). The challenge is determining how many such groups are necessary given the frequency count.
   • Use the formula ((count + answer) / (answer + 1)) to compute how many full groups are needed, making sure to round up. This is achieved using integer division: count + answer ensures that any remainder will force an additional group being included due to integer rounding.
   • Multiply the number of groups required by the group size answer + 1 to ascertain the total number of rabbits for that particular answer category.

3. Sum Up the Total:

   • Accumulate the total rabbits calculated for each distinct answer to get the minimum number of rabbits in the forest.

By following this approach, we ensure that all rabbits accounted for both those who answered and those logically inferred from the answers provided. The final count from these calculations yields the desired result.

Code

C++

class Solution {
public:
    int numRabbits(vector<int>& answers) {
        int totalRabbits = 0; // Initialize the total number of rabbits.
        map<int, int> countMap; // Map to track the count of each unique answer.

        // Count the frequency of each distinct answer.
        for (int answer : answers) {
            countMap[answer]++;
        }

        // Calculate the minimum number of rabbits based on the collected answers.
        for (auto& [answer, count] : countMap) {
            // 'answer + 1' represents the group size (same colored rabbits).
            // Calculate the number of such groups needed using integer division rounding up.
            // Multiply by the group size to get the total number of rabbits for this answer.
            totalRabbits += (answer + 1) * ((count + answer) / (answer + 1));
        }

        return totalRabbits;
    }
};

Python

class Solution:
    def numRabbits(self, answers):
        totalRabbits = 0  # Initialize the total number of rabbits.
        countMap = {}  # Dictionary to track the count of each unique answer.

        # Count the frequency of each distinct answer.
        for answer in answers:
            if answer in countMap:
                countMap[answer] += 1
            else:
                countMap[answer] = 1

        # Calculate the minimum number of rabbits based on the collected answers.
        for answer, count in countMap.items():
            # 'answer + 1' represents the group size (same colored rabbits).
            # Calculate the number of such groups needed using integer division rounding up.
            # Multiply by the group size to get the total number of rabbits for this answer.
            totalRabbits += (answer + 1) * ((count + answer) // (answer + 1))

        return totalRabbits

Complexity

• Time complexity: $O(n)$

  The solution primarily involves two loops: one just iterates over the input array answers to populate the countMap, and the other iterates through the unique keys in countMap. The first loop takes $O(n)$ time where $n$ is the length of the answers array. The second loop iterates over the distinct answers, each of which occurs at least once in answers. In the worst case, this loop could take $O(n)$ time if all elements are distinct. Thus, the overall time complexity is $O(n)$.

• Space complexity: $O(n)$

  The space used by the algorithm is dominated by the countMap, which stores a mapping of answers to their frequencies. In the worst case, each answer in the answers array is unique, leading to an additional space requirement of $O(n)$ to store these entries in the map. Therefore, the space complexity is $O(n)$.