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Leetcode 781. Rabbits in Forest

Problem Statement

The problem is to count the minimum number of rabbits that could be in the forest based on their answers. Each rabbit must answer “Y + 1” where ‘Y’ is the number of other rabbits that have the same color which it has.

Given an array of integers, where each integer answers[i] represents the answer a rabbit gave, calculate the minimum number of rabbits that could be in the forest.

Example:

Input: answers = [1, 1, 2]
Output: 5
Explanation: 
- The two rabbits that answered "1" could both be of one color.
- The rabbit that answered "2" can't be the same color as the rabbits above. Hence, there would be at least 3 rabbits answering "2".

Clarifying Questions

  1. What is the range of the input values?
    • The values in the answers array can be between 0 and 999. The length of the array can be between 1 and 1000.
  2. Can the answers array be empty?
    • No, based on the constraints the array will have at least one element.
  3. What should be returned in the case of an empty input array?
    • This is not applicable as per the problem constraints.

Strategy

  1. Understanding the Problem:
    • Each answers[i] gives the number of other rabbits with the same color. Therefore, if a rabbit says answers[i] = 2, it means there are 2 other rabbits of the same color as it, totaling 3 rabbits with that color.
  2. Group Rabbits Based on Answers:
    • For each unique answer, count how many rabbits gave that answer.
    • From these counts, calculate the minimum number of groups required. Each group will have answers[i] + 1 rabbits.
  3. Calculate the Total Number of Rabbits:
    • For each unique answer, if n rabbits say answers[i], the minimum number of groups of rabbits of that color will be (n // (answers[i] + 1)) + (n % (answers[i] + 1) != 0). The total number of rabbits for that answer will be the number of groups multiplied by answers[i] + 1.

Code

Here’s the solution implemented in C++:

#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 answer : answers) {
            count[answer]++;
        }
        
        int result = 0;
        for (auto& [answer, num_rabbits] : count) {
            int group_size = answer + 1;
            int groups = ceil(static_cast<double>(num_rabbits) / group_size);
            result += groups * group_size;
        }
        
        return result;
    }
};

Time Complexity

The time complexity of this solution is O(N), where N is the number of elements in the answers array. This is because we go through the answers list once to create the count map and then iterate through the counts of each unique answer.

The space complexity is also O(N) due to the unordered_map used to store the count of each answer.

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