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 <bits/stdc++.h>
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 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 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# 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
<script>
// 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);
</script>
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 <bits/stdc++.h>
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[0]);
int k = 2;
if (checkKTimesElement(arr, n, k))
cout << "Yes\n";
else
cout << "No\n";
return 0;
}
// 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<Integer> s = new HashSet<Integer>();
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 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# 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
<script>
//Javascript 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
function checkKTimesElement(arr, n, k)
{
// Create an empty set
s = new Set();
for (var i = 0; i < n; i++){
s.add(arr[i]);
}
for (var i = 0; i < n; i++) {
// Check if K times of
// element exists in set
if (s.has(arr[i] * k))
return true;
}
return false;
}
// Driver program to test above
var arr = [5, 14, 8, 13, 10];
var n = arr.length;
var k = 2;
if (checkKTimesElement(arr, n, k))
document.write("Yes");
else
document.write("No");
// This code is contributed by shivani.
</script>
Time Complexity: O(n)
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