当一个人不能携带超过两件相同类型的物品时分发物品

给定 N 个糖果，它可以有多种不同类型和 k 个顾客，一个顾客不会制作超过 2 件的同一种糖果，任务是寻找是否可以分发所有糖果，然后打印“是”或否则“否”。
给定一个数组，arr[] 表示一个糖果数组。 arr[i] 是一种 Sweet 。

例子：

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 : Yes

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

方法一：
1- 遍历每个元素的数组。
2- 计算数组中每个元素的出现次数
3- 检查每个元素的结果是否必须小于或等于 2*k。

C++

// C++ program for above implementation
#include 
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

 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.
?>

Javascript

    // Javascript program for above implementation
     
    // Function to check occurrence
    // of each element
    function checkCount(arr, n, k)
    {
        let count;
      
        // Start traversing the elements
        for (let i = 0; i < n; i++)
        {
      
            // Count occurrences of
            // current element
            count = 0;
            for (let 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;
    }
     
    let arr = [ 1, 1, 2, 3, 1 ];
    let n = arr.length;
    let k = 2;
 
    if(checkCount(arr, n, k))
      document.write("Yes");
    else
      document.write("No");
             

输出：

Yes

时间复杂度： O(n^2)

方法二：
1. 维护 32 种不同类型糖果的哈希值。
2.遍历一个数组并检查每个arr[i]

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

C++

// C++ program for above implementation
#include 
using namespace std;
 
// Function to check hash array
// corresponding to the given array
bool checkCount(int arr[], int n, int k)
{
    unordered_map 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  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

蟒蛇3

# 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 hash = new Dictionary();
     
        // 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 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 */

输出：

Yes

时间复杂度： O(n)

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