检查数组中是否存在元素的K次

给定数组arr []和整数K ，任务是检查数组中是否还存在K次任何元素。
例子 ：

Input: arr[] = {10, 14, 8, 13, 5}, K = 2 
Output: Yes 
Explanation: 
K times of 5 is also present in array, i.e. 10.
Input: arr[] = {7, 8, 5, 9, 11}, K = 3 
Output: No 
Explanation: 
K times of any element is not present in the array 

天真的方法：一个简单的解决方案是运行两个嵌套循环，并检查数组中是否也存在该元素的K倍的每个元素。
下面是上述方法的实现：

C++

// C++ implementation to check whether
// K times of a element is present in
// the array
 
#include 
 
using namespace std;
 
// Function to find if K times of
// an element exists in array
void checkKTimesElement(int arr[], int n, int k)
{
    bool found = false;
     
    // Loop to check that K times of
    // element is present in the array
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (arr[j] == k * arr[i]) {
                found = true;
                break;
            }
        }
    }
 
    if (found)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
 
// Driver code
int main()
{
    int arr[] = { 10, 14, 8, 13, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
     
    // Function Call
    checkKTimesElement(arr, n, k);
    return 0;
}

Java

// Java implementation to check whether
// K times of a element is present in
// the array
 
class GFG{
 
// Function to find if K times of
// an element exists in array
static void checkKTimesElement(int arr[],
                               int n,
                               int k)
{
    boolean found = false;
     
    // Loop to check that K times of
    // element is present in the array
    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j < n; j++)
       {
          if (arr[j] == k * arr[i])
          {
              found = true;
              break;
          }
       }
    }
 
    if (found)
        System.out.print("Yes" + "\n");
    else
        System.out.print("No" + "\n");
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 10, 14, 8, 13, 5 };
    int n = arr.length;
    int k = 2;
     
    // Function call
    checkKTimesElement(arr, n, k);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to check whether
# K times of a element is present in
# the array
 
# Function to find if K times of
# an element exists in array
def checkKTimesElement(arr, n, k):
     
    found = False
     
    # Loop to check that K times of
    # element is present in the array
    for i in range(0, n):
        for j in range(0, n):
            
            if arr[j] == k * arr[i]:
                found = True
                break
    if found:
        print('Yes')
    else:
        print('No')
         
# Driver code
if __name__=='__main__':
     
    arr = [ 10, 14, 8, 13, 5 ]
    n = len(arr)
    k = 2
     
    # Function Call
    checkKTimesElement(arr, n, k)
 
# This code is contributed by rutvik_56

C#

// C# implementation to check whether
// K times of a element is present in
// the array
using System;
 
class GFG{
 
// Function to find if K times of
// an element exists in array
static void checkKTimesElement(int []arr,
                               int n,
                               int k)
{
    bool found = false;
     
    // Loop to check that K times of
    // element is present in the array
    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j < n; j++)
       {
          if (arr[j] == k * arr[i])
          {
              found = true;
              break;
          }
       }
    }
    if (found)
        Console.Write("Yes" + "\n");
    else
        Console.Write("No" + "\n");
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 10, 14, 8, 13, 5 };
    int n = arr.Length;
    int k = 2;
     
    // Function call
    checkKTimesElement(arr, n, k);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

 
// Javascript implementation to check whether
// K times of a element is present in
// the array
 
// Function to find if K times of
// an element exists in array
function checkKTimesElement(arr, n, k)
{
    var found = false;
     
    // Loop to check that K times of
    // element is present in the array
    for (var i = 0; i < n; i++) {
        for (var j = 0; j < n; j++) {
            if (arr[j] == k * arr[i]) {
                found = true;
                break;
            }
        }
    }
 
    if (found)
        document.write( "Yes");
    else
        document.write( "No" );
}
 
// Driver code
var arr = [ 10, 14, 8, 13, 5 ];
var n = arr.length;
var k = 2;
 
// Function Call
checkKTimesElement(arr, n, k);
 

输出：

Yes

高效的方法：该想法是将所有元素存储在哈希图中，并针对每个元素检查该哈希图中是否存在该元素的K次。如果它存在于哈希图中，则返回True，否则返回False。
下面是上述方法的实现：

C++

// C++ implementation to check whether
// K times of a element is present in
// the array
 
#include 
 
using namespace std;
 
// Function to check if K times of
// an element exists in array
bool checkKTimesElement(int arr[], int n, int k)
{
    // Create an empty set
    unordered_set s;
    for (int i = 0; i < n; i++){
        s.insert(arr[i]);
    }
 
    for (int i = 0; i < n; i++) {
         
        // Check if K times of
        // element exists in set
        if (s.find(arr[i] * k) != s.end())
            return true;
    }
 
    return false;
}
 
// Driven code
int main()
{
    int arr[] = { 5, 14, 8, 13, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
     
    if (checkKTimesElement(arr, n, k))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

Java

// Java implementation to check whether
// K times of a element is present in
// the array
import java.util.*;
 
class GFG{
 
// Function to check if K times of
// an element exists in array
static boolean checkKTimesElement(int arr[], int n,
                                             int k)
{
     
    // Create an empty set
    HashSet s = new HashSet();
     
    for(int i = 0; i < n; i++)
    {
       s.add(arr[i]);
    }
 
    for(int i = 0; i < n; i++)
    {
         
       // Check if K times of
       // element exists in set
       if (s.contains(arr[i] * k))
           return true;
    }
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 14, 8, 13, 10 };
    int n = arr.length;
    int k = 2;
     
    if (checkKTimesElement(arr, n, k))
        System.out.print("Yes\n");
    else
        System.out.print("No\n");
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to
# check whether K times of 
# a element is present in the array
  
# Function to check if K times of
# an element exists in array
def checkKTimesElement(arr, n, k):
 
  # Create an empty set
  s = set([])
  for i in range (n):
    s.add(arr[i])
  for i in range (n):
 
    # Check if K times of
    # element exists in set
    if ((arr[i] * k) in s):
      return True
    return False
  
# Driver code
if __name__ == "__main__":
 
  arr = [5, 14, 8, 13, 10]
  n = len(arr)
  k = 2
 
  if (checkKTimesElement(arr, n, k)):
    print ("Yes")
  else:
    print("No")
 
# This code is contributed by Chitranayal

C＃

// C# implementation to check whether
// K times of a element is present in
// the array
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if K times of
// an element exists in array
static bool checkKTimesElement(int[] arr, int n,
                                          int k)
{
         
    // Create an empty set
    HashSet s = new HashSet();
         
    for(int i = 0; i < n; i++)
    {
       s.Add(arr[i]);
    }
     
    for(int i = 0; i < n; i++)
    {
        
       // Check if K times of
       // element exists in set
       if (s.Contains(arr[i] * k))
           return true;
    }
    return false;
}
     
// Driver code
static public void Main ()
{
    int[] arr = { 5, 14, 8, 13, 10 };
    int n = arr.Length;
    int k = 2;
         
    if (checkKTimesElement(arr, n, k))
        Console.Write("Yes\n");
    else
        Console.Write("No\n");
}
}
 
// This code is contributed by ShubhamCoder

输出：

Yes

时间复杂度： O(n)