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    Minimum number of rabbits that must be present in the forest
    • Difficulty Level : Medium
    • Last Updated : 07 Dec, 2020

    There are some colored rabbits in a forest. Given an array arr[] of size N, such that arr[i] denotes the number of rabbits having same color as the ith rabbit, the task is to find the minimum number of rabbits that could be in the forest.

    Examples:

    Input: arr[] = {2, 2, 0}
    Output: 4
    Explanation: Considering the 1st and the 2nd rabbits to be of same color, eg. Blue, there should be 3 blue-colored rabbits. The third rabbit is the only rabbit of that color. Therefore, the minimum number of rabbits that could be present in the forest are = 3 + 1 = 4.

    Input: arr[] = {10, 10, 10}
    Output: 11
    Explanation: Considering all the rabbits to be of the same color, the minimum number of rabbits present in forest are 10 + 1 = 11.

    Approach: The approach to solving this problem is to find the number of groups of rabbits that have the same color and the number of rabbits in each group. Below are the steps:



    • Initialize a variable count to store the number of rabbits in each group.
    • Initialize a map and traverse the array having key as arr[i] and value as occurrences of arr[i] in the given array.
    • Now, if y rabbits answered x, then:
      • If (y%(x + 1)) is 0, then there must be (y / (x + 1)) groups of (x + 1) rabbits.
      • If (y % (x + 1)) is non-zero, then there must be (y / (x + 1)) + 1 groups of (x + 1) rabbits.
    • Add the product of the number of groups and the number of rabbits in each group to the variable count.
    • After the above steps, the value of count gives the minimum number of rabbits in the forest.

    Below is the implementation of the above approach:

    C++

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    // C++ program for the above approach
    #include <bits/stdc++.h>
    using namespace std;
     
    // Function to find the minimun
    // number of rabbits in the forest
    int minNumberOfRabbits(int answers[], int N)
    {
         
        // Initialize map
        map<int, int> Map;
         
        // Traverse array and map arr[i]
        // to the number of occurences
        for(int a = 0; a < N; a++)
        {
            Map[answers[a]]++;
        }
     
        // Intialize count as 0;
        int count = 0;
     
        // Find the number groups and
        // no. of rabbits in each group
        for(auto a : Map)
        {
            int x = a.first;
            int y = a.second;
             
            // Find number of groups and
            // multiply them with number
            // of rabbits in each group
            if (y % (x + 1) == 0) 
                count = count + (y / (x + 1)) *
                                     (x + 1);
            else
                count = count + ((y / (x + 1)) + 1) *
                                      (x + 1);
        }
     
        // count gives minimum number
        // of rabbits in the forest
        return count;
    }
     
    // Driver code   
    int main()
    {
        int arr[] = { 2, 2, 0 };
        int N = sizeof(arr) / sizeof(arr[0]);
     
        // Function Call
        cout << minNumberOfRabbits(arr, N) << endl;
     
        return 0;
    }
     
    // This code is contributed by divyeshrabadiya07

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    Java

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    // Java program for the above approach
     
    import java.util.*;
     
    class GFG {
     
        // Function to find the minimun
        // number of rabbits in the forest
        public static int minNumberOfRabbits(
            int[] answers, int N)
        {
            // Initialize map
            Map<Integer, Integer> map
                = new HashMap<Integer, Integer>();
     
            // Traverse array and map arr[i]
            // to the number of occurences
            for (int a : answers) {
                map.put(a, map.getOrDefault(a, 0) + 1);
            }
     
            // Intialize count as 0;
            int count = 0;
     
            // Find the number groups and
            // no. of rabbits in each group
            for (int a : map.keySet()) {
                int x = a;
                int y = map.get(a);
     
                // Find number of groups and
                // multiply them with number
                // of rabbits in each group
                if (y % (x + 1) == 0) {
                    count = count + (y / (x + 1)) * (x + 1);
                }
                else
                    count
                        = count + ((y / (x + 1)) + 1) * (x + 1);
            }
     
            // count gives minimum number
            // of rabbits in the forest
            return count;
        }
     
        // Driver Code
        public static void main(String[] args)
        {
            int arr[] = { 2, 2, 0 };
            int N = arr.length;
     
            // Function Call
            System.out.println(
                minNumberOfRabbits(arr, N));
        }
    }

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    Python3

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    # Python3 program for the above approach
     
    # Function to find the minimun
    # number of rabbits in the forest
    def minNumberOfRabbits(answers, N) :
         
        # Initialize map
        Map = {}
         
        # Traverse array and map arr[i]
        # to the number of occurences
        for a in range(N) :
             
            if answers[a] in Map :
                Map[answers[a]] += 1
            else :
                Map[answers[a]] = 1
                 
        # Intialize count as 0;
        count = 0
     
        # Find the number groups and
        # no. of rabbits in each group
        for a in Map :
         
            x = a;
            y = Map[a]
             
            # Find number of groups and
            # multiply them with number
            # of rabbits in each group
            if (y % (x + 1) == 0) :
                count = count + (y // (x + 1)) * (x + 1)
            else :
                count = count + ((y // (x + 1)) + 1) * (x + 1)
     
        # count gives minimum number
        # of rabbits in the forest
        return count
     
    # Driver code 
    arr = [ 2, 2, 0 ]
    N = len(arr)
     
    # Function Call
    print(minNumberOfRabbits(arr, N))
     
    # This code is contributed by divyesh072019

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    C#

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    // C# program for the above approach
    using System;
    using System.Collections.Generic;
    using System.Linq;
     
    class GFG{
     
    // Function to find the minimun
    // number of rabbits in the forest
    public static int minNumberOfRabbits(int[] answers,
                                         int N)
    {
         
        // Initialize map
        Dictionary<int,
                   int> map = new Dictionary<int,
                                             int>();
     
        // Traverse array and map arr[i]
        // to the number of occurences
        for(int a = 0; a < N; a++)
        {
            if (map.ContainsKey(answers[a]))
                map[answers[a]] += 1;
            else
                map.Add(answers[a], 1);
        }
     
        // Intialize count as 0;
        int count = 0;
     
        // Find the number groups and
        // no. of rabbits in each group
        for(int a = 0; a < map.Count; a++)
        {
            int x = map.Keys.ElementAt(a);
            int y = map[x];
     
            // Find number of groups and
            // multiply them with number
            // of rabbits in each group
            if (y % (x + 1) == 0)
            {
                count = count + (y / (x + 1)) * (x + 1);
            }
            else
                count = count + ((y / (x + 1)) + 1) *
                           (x + 1);
        }
     
        // count gives minimum number
        // of rabbits in the forest
        return count;
    }
     
    // Driver Code
    public static void Main(string[] args)
    {
        int []arr = { 2, 2, 0 };
        int N = arr.Length;
     
        // Function Call
        Console.WriteLine(
            minNumberOfRabbits(arr, N));
    }
    }
     
    // This code is contributed by AnkitRai01

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    Output: 

    4

     

    Time Complexity: O(N)
    Auxiliary Space: O(N)

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