Distributing items when a person cannot take more than two items of same type

Given N sweets which can be of many different types and k customers, one customer won’t take the same type of sweet more than 2 pieces, the task is to find if it is possible to distribute all sweets then print “Yes” otherwise “No”.

Given array arr[] represents array of sweets arr[i] is type of sweet.

Examples:

Input : arr[] = {1, 1, 2, 3, 1}, 
            k = 2;
Output : Yes
There are three pieces of sweet type 1,
one piece of type 2 and one piece of 
type 3. Two customers can distribute 
sweets under given constraints.

Input : arr[] = {2, 3, 3, 5, 3, 3}, 
            k = 2;
Output : No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1:
1- Traverse array for each element.
2- Count occurrences of each element in array
3- Check Resulting occurrence of each element must be less than or equal to 2*k.

C++

// C++ program for above implementation 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check occurrence of each element 
bool checkCount(int arr[], int n, int k) 
{ 
    int count; 
  
    // Start traversing the elements 
    for (int i = 0; i < n; i++) { 
  
        // Count occurrences of current element 
        count = 0; 
        for (int j = 0; j < n; j++) { 
            if (arr[j] == arr[i]) 
                count++; 
  
            // If count of any element is greater 
            // than 2*k then return false 
            if (count > 2 * k) 
                return false; 
        } 
    } 
  
    return true; 
} 
  
// Drivers code 
int main() 
{ 
    int arr[] = { 1, 1, 2, 3, 1 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    int k = 2; 
    checkCount(arr, n, k) ? cout << "Yes"
                           : cout << "No"; 
  
    return 0; 
} 

Java

// java program for above implementation 
import java.io.*; 
  
public class GFG { 
      
    // Function to check occurrence of 
    // each element 
    static boolean checkCount(int []arr, 
                            int n, int k) 
    { 
        int count; 
      
        // Start traversing the elements 
        for (int i = 0; i < n; i++) 
        { 
      
            // Count occurrences of 
            // current element 
            count = 0; 
            for (int j = 0; j < n; j++) 
            { 
                if (arr[j] == arr[i]) 
                    count++; 
      
                // If count of any element 
                // is greater than 2*k then 
                // return false 
                if (count > 2 * k) 
                    return false; 
            } 
        } 
      
        return true; 
    } 
      
    // Drivers code 
    static public void main (String[] args) 
    { 
        int []arr = { 1, 1, 2, 3, 1 }; 
        int n = arr.length; 
        int k = 2; 
          
        if(checkCount(arr, n, k))  
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
} 
  
// This code is contributed by vt_m. 

Python3

# Python 3 program for above implementation  
  
# Function to check occurrence  
# of each element  
def checkCount(arr, n, k): 
  
    # Start traversing the elements  
    for i in range(n): 
  
        # Count occurrences of  
        # current element  
        count = 0
        for j in range(n): 
            if arr[j] == arr[i]: 
                count += 1
  
            # If count of any element is greater  
            # than 2*k then return false  
            if count > 2 * k: 
                return False
    return True
  
# Driver code 
arr = [1, 1, 2, 3, 1] 
n = len(arr) 
k = 2
if checkCount(arr, n, k) == True: 
    print("Yes") 
else: 
    print("No") 
  
# This code is contributed by Shrikant13 

C#

// C# program for above implementation 
using System; 
  
public class GFG { 
      
    // Function to check occurrence 
    // of each element 
    static bool checkCount(int []arr, 
                          int n, int k) 
    { 
        int count; 
      
        // Start traversing the elements 
        for (int i = 0; i < n; i++) 
        { 
      
            // Count occurrences of 
            // current element 
            count = 0; 
            for (int j = 0; j < n; j++) 
            { 
                if (arr[j] == arr[i]) 
                    count++; 
      
                // If count of any element  
                // is greater than 2*k then 
                // return false 
                if (count > 2 * k) 
                    return false; 
            } 
        } 
      
        return true; 
    } 
      
    // Drivers code 
    static public void Main () 
    { 
        int []arr = { 1, 1, 2, 3, 1 }; 
        int n = arr.Length; 
        int k = 2; 
          
        if(checkCount(arr, n, k))  
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program for above implementation 
  
// Function to check occurrence 
// of each element 
function checkCount($arr, $n, $k) 
{ 
    $count; 
  
    // Start traversing the elements 
    for($i = 0; $i < $n; $i++)  
    { 
  
        // Count occurrences of  
        // current element 
        $count = 0; 
        for($j = 0; $j < $n; $j++) 
        { 
            if ($arr[$j] == $arr[$i]) 
                $count++; 
  
            // If count of any element  
            // is greater than 2*k then 
            // return false 
            if ($count > 2 * $k) 
                return false; 
        } 
    } 
  
    return true; 
} 
  
    // Driver Code 
    $arr = array(1, 1, 2, 3, 1); 
    $n =count($arr); 
    $k = 2; 
    if(checkCount($arr, $n, $k)) 
        echo "Yes"; 
    else
        echo "No"; 
  
// This code is contributed by anuj_67. 
?> 

Output:

Yes

Time Complexity: O(n^2)

Method 2:
1. Maintain a hash for 32 different type of sweets.
2. Traverse an array and check for every arr[i]

hash[arr[i]] <= 2*k.

C++

// C++ program for above implementation 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check hash array 
// corresponding to the given array 
bool checkCount(int arr[], int n, int k) 
{ 
    unordered_map<int, int> hash; 
  
    // Maintain a hash 
    for (int i = 0; i < n; i++) 
        hash[arr[i]]++; 
  
    // Check for each value in hash 
    for (auto x : hash) 
        if (x.second > 2 * k) 
            return false; 
  
    return true; 
} 
  
// Drivers code 
int main() 
{ 
    int arr[] = { 1, 1, 2, 3, 1 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    int k = 2; 
    checkCount(arr, n, k) ? cout << "Yes"
                           : cout << "No"; 
    return 0; 
} 

Java

// Java program for above implementation 
import java.util.HashMap; 
import java.util.Map; 
  
class GfG 
{ 
  
    // Function to check hash array  
    // corresponding to the given array  
    static boolean checkCount(int arr[], int n, int k)  
    {  
        HashMap <Integer, Integer> hash = new HashMap<>();  
      
        // Maintain a hash  
        for (int i = 0; i < n; i++) 
        { 
            if (!hash.containsKey(arr[i])) 
                hash.put(arr[i], 0); 
            hash.put(arr[i], hash.get(arr[i]) + 1); 
        } 
      
        // Check for each value in hash  
        for (Map.Entry x : hash.entrySet()) 
            if ((int)x.getValue() > 2 * k)  
                return false;  
      
        return true;  
    }  
  
    // Driver code 
    public static void main(String []args) 
    { 
          
        int arr[] = { 1, 1, 2, 3, 1 };  
        int n = arr.length;  
        int k = 2;  
        if (checkCount(arr, n, k))  
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
} 
  
// This code is contributed by Rituraj Jain 

Python3

# Python3 program for above implementation  
from collections import defaultdict 
  
# Function to check hash array  
# corresponding to the given array  
def checkCount(arr, n, k): 
  
    mp = defaultdict(lambda:0) 
  
    # Insert all array elements in 
    # hash table Maintain a hash 
    for i in range(n): 
        mp[arr[i]] += 1
  
    # Check for each value in hash 
    for key, values in mp.items(): 
        if values > 2 * k: 
            return False
    return True
  
# Driver code 
arr = [ 1, 1, 2, 3, 1 ] 
n = len(arr) 
k = 2
if checkCount(arr, n, k) == True: 
    print("Yes") 
else: 
    print("No") 
  
# This code is contributed by Shrikant13  

C#

// C# program for above implementation 
using System;  
using System.Collections.Generic; 
  
class GfG 
{ 
  
    // Function to check hash array  
    // corresponding to the given array  
    static Boolean checkCount(int []arr, int n, int k)  
    {  
        Dictionary<int,int> hash = new Dictionary<int,int>();  
      
        // Maintain a hash  
        for (int i = 0; i < n; i++) 
        { 
            if(hash.ContainsKey(arr[i])) 
            { 
                var val = hash[arr[i]]; 
                hash.Remove(arr[i]); 
                hash.Add(arr[i], val + 1);  
            } 
            else
            { 
                hash.Add(arr[i], 0); 
            } 
        } 
      
        // Check for each value in hash  
        foreach(KeyValuePair<int, int> x in hash) 
            if ((int)x.Value > 2 * k)  
                return false;  
      
        return true;  
    }  
  
    // Driver code 
    public static void Main(String []args) 
    { 
          
        int []arr = { 1, 1, 2, 3, 1 };  
        int n = arr.Length;  
        int k = 2;  
        if (checkCount(arr, n, k))  
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
  
/* This code is contributed by PrinciRaj1992 */

Output:

Yes

Time Complexity: O(n)

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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