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# Check whether K times of a element is present in array

• Last Updated : 24 May, 2021

Given an array arr[] and an integer K, the task is to check whether K times of any element are also present in the array.

Examples :

Input: arr[] = {10, 14, 8, 13, 5}, K = 2
Output: Yes
Explanation:
K times of 5 is also present in an 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

Naive Approach: A simple solution is to run two nested loops and check for every element that K times of that element is also present in the array.

Below is the implementation of the above approach:

## 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);``    ``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

 ``
Output:
`Yes`

Efficient Approach: The idea is to store all elements in hash-map and for each element check that K times of that element is present in the hash-map. If it exists in the hash-map, then return True otherwise False.

Below is the implementation of the above approach:

## 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<``int``> 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);``    ``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<``int``> s = ``new` `HashSet<``int``>();``        ` `    ``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`

## Javascript

 ``
Output:
`Yes`

Time Complexity: O(n)

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