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++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum // 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 occurrences
for ( int a = 0; a < N; a++) {
Map[answers[a]]++;
}
// Initialize 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 |
// Java program for the above approach import java.util.*;
class GFG {
// Function to find the minimum
// 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 occurrences
for ( int a : answers) {
map.put(a, map.getOrDefault(a, 0 ) + 1 );
}
// Initialize 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));
}
} |
# Python3 program for the above approach # Function to find the minimum # number of rabbits in the forest def minNumberOfRabbits(answers, N):
# Initialize map
Map = {}
# Traverse array and map arr[i]
# to the number of occurrences
for a in range (N):
if answers[a] in Map :
Map [answers[a]] + = 1
else :
Map [answers[a]] = 1
# Initialize 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 |
// C# program for the above approach using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
// Function to find the minimum
// 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 occurrences
for ( int a = 0; a < N; a++) {
if (map.ContainsKey(answers[a]))
map[answers[a]] += 1;
else
map.Add(answers[a], 1);
}
// Initialize 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 |
<script> // JavaScript program for the above approach // Function to find the minimum // number of rabbits in the forest function minNumberOfRabbits(answers, N)
{ // Initialize map
var map = new Map();
// Traverse array and map arr[i]
// to the number of occurrences
for ( var a = 0; a < N; a++) {
if (map.has(answers[a]))
map.set(answers[a], map.get(answers[a])+1)
else
map.set(answers[a], 1)
}
// Initialize count as 0;
var count = 0;
// Find the number groups and
// no. of rabbits in each group
map.forEach((value, key) => {
var x = key;
var y = value;
// Find number of groups and
// multiply them with number
// of rabbits in each group
if (y % (x + 1) == 0)
count = count + parseInt(y / (x + 1)) * (x + 1);
else
count = count + (parseInt(y / (x + 1)) + 1) * (x + 1);
});
// count gives minimum number
// of rabbits in the forest
return count;
} // Driver code var arr = [2, 2, 0];
var N = arr.length;
// Function Call document.write( minNumberOfRabbits(arr, N)); </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(N)
Another solution is to use HashMap of the counts and decrease the count every time the same answers[i] is met. If the same answers[i] occur again and the count is zero, reset the count with answers[i]+1 and add to the answer variable. Below are the steps:
- Initialize a variable count with zero.
- Use an unordered map mp.
- Traverse the array and do the following:
- if answers[i] is set to zero, set mp[answers[i]] = answers[i]+1 and count=count+answers[i]+1
- finally subtract the count from map, mp[answers[i]]–;
- return the cnt as answer.
// C++ program for the above approach #include <iostream> #include <unordered_map> using namespace std;
// Function to find the minimum // number of rabbits in the forest int minNumberOfRabbits( int answers[], int n)
{ // Initialize cnt variable
int count = 0;
// Initialize map
unordered_map< int , int > mp;
for ( int i = 0; i < n; ++i) {
// Add to the count if not found or
// has exhausted the count of all the
// number of rabbits of same colour
if (mp[answers[i]] == 0) {
count += answers[i] + 1;
mp[answers[i]] = answers[i] + 1;
}
mp[answers[i]]--;
}
// count gives minimum number
// of rabbits in the forest
return count;
} // Driver Code int main()
{ int arr[] = { 10, 10, 0 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
cout << minNumberOfRabbits(arr, N) << endl;
return 0;
} // This code is contributed by Bhavna Soni - bhavna23 |
// Java program for the above approach import java.util.*;
class GFG {
// Function to find the minimum
// number of rabbits in the forest
static int minNumberOfRabbits( int [] answers, int n)
{
// Initialize cnt variable
int count = 0 ;
// Initialize map
HashMap<Integer, Integer> mp = new HashMap<Integer,Integer>();
for ( int i = 0 ; i < n; ++i) {
// Add to the count if not found or
// has exhausted the count of all the
// number of rabbits of same colour
if (!mp.containsKey(answers[i]))
{
count += answers[i] + 1 ;
mp.put(answers[i],answers[i]+ 1 );
}
if (mp.containsKey(answers[i]))
mp.put(answers[i],mp.get(answers[i])- 1 );
}
// count gives minimum number
// of rabbits in the forest
return count;
} // Driver Code
public static void main(String[] args)
{
int arr[] = { 10 , 10 , 0 };
int N = arr.length;
// Function Call
System.out.println(minNumberOfRabbits(arr, N));
}
} // This code is contributed by Pushpesh Raj |
# Python 3 program for the above approach # Function to find the minimum # number of rabbits in the forest def minNumberOfRabbits(answers, n):
# Initialize cnt variable
count = 0
# Initialize map
mp = {}
for i in range (n):
# Add to the count if not found or
# has exhausted the count of all the
# number of rabbits of same colour
if (answers[i] not in mp):
count + = answers[i] + 1
mp[answers[i]] = answers[i] + 1
mp[answers[i]] - = 1
# count gives minimum number
# of rabbits in the forest
return count
# Driver Code if __name__ = = "__main__" :
arr = [ 10 , 10 , 0 ]
N = len (arr)
# Function Call
print (minNumberOfRabbits(arr, N))
# This code is contributed by ukasp.
|
// Include namespace system using System;
using System.Collections.Generic;
using System.Collections;
public class GFG
{ // Function to find the minimum
// number of rabbits in the forest
public static int minNumberOfRabbits( int [] answers, int n)
{
// Initialize cnt variable
var count = 0;
// Initialize map
var mp = new Dictionary< int , int >();
for ( int i = 0; i < n; ++i)
{
// Add to the count if not found or
// has exhausted the count of all the
// number of rabbits of same colour
if (!mp.ContainsKey(answers[i]))
{
count += answers[i] + 1;
mp[answers[i]] = answers[i] + 1;
}
if (mp.ContainsKey(answers[i]))
{
mp[answers[i]] = mp[answers[i]] - 1;
}
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver Code
public static void Main(String[] args)
{
int [] arr = {10, 10, 0};
var N = arr.Length;
// Function Call
Console.WriteLine(GFG.minNumberOfRabbits(arr, N));
}
} // This code is contributed by sourabhdalal0001. |
<script> class GFG { // Function to find the minimum
// number of rabbits in the forest
static minNumberOfRabbits(answers, n)
{
// Initialize cnt variable
var count = 0;
// Initialize map
var mp = new Map();
for ( var i=0; i < n; ++i)
{
// Add to the count if not found or
// has exhausted the count of all the
// number of rabbits of same colour
if (!mp.has(answers[i]))
{
count += answers[i] + 1;
mp.set(answers[i],answers[i] + 1);
}
if (mp.has(answers[i]))
{
mp.set(answers[i],mp.get(answers[i]) - 1);
}
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver Code
static main(args)
{
var arr = [10, 10, 0];
var N = arr.length;
// Function Call
document.write(GFG.minNumberOfRabbits(arr, N));
}
} GFG.main([]); </script> |
12
Time Complexity: O(N)
Auxiliary Space: O(N)