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Find the all elements that appears n times

Here given code implementation process.

// C Program
// Find the all elements that appears n times
#include <stdio.h>

//Function which is display array elements
void display(int arr[], int size)
{
	for (int i = 0; i < size; ++i)
	{
		printf("  %d", arr[i]);
	}
}
// Display the all elements which is occurs of n times in given collection
void find_element(int arr[], int n, int size)
{
	if (n <= 0)
	{
		return;
	}
	// Display element
	display(arr, size);
	// Loop controlling variables
	int i;
	int j;
	// Define counter variable
	int counter = 0;
	int status = 0;
	printf("\n Occurring elements of %d times is : ", n);
	// Outer loop
	for (i = 0; i < size; ++i)
	{
		// Inner loop
		for (j = 0; j < size && counter >= 0 && counter <= n; ++j)
		{
			if (arr[i] == arr[j])
			{
				if (j < i)
				{
					// If element has been already tested
					counter = -1;
				}
				else
				{
					counter++;
				}
			}
		}
		if (counter == n)
		{   // When get new element
			status = 1;
			printf("  %d", arr[i]);
		}
		counter = 0;
	}
	if (status == 0)
	{
		printf("None");
	}
	printf("\n\n");
}
int main()
{
	int arr[] = {
		2 , 7 , 1 , 8 , 3 , 0 , 4 , 1 , 4 , 3 , 6 , 2 , 5 , 6 , 1 , 2 , 3 , 6
	};
	int size = sizeof(arr) / sizeof(arr[0]);
	int n = 3;
	find_element(arr, n, size);
	return 0;
}

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

/*
  Java Program
  Find the all elements that appears n times
*/
public class SearchElement
{
	//Function which is display array elements
	public void display(int[] arr, int size)
	{
		for (int i = 0; i < size; ++i)
		{
			System.out.print("  " + arr[i]);
		}
	}
	// Display the all elements which is occurs of n times in given collection
	public void find_element(int[] arr, int n, int size)
	{
		if (n <= 0)
		{
			return;
		}
		// Display element
		display(arr, size);
		// Loop controlling variables
		int i;
		int j;
		// Define counter variable
		int counter = 0;
		boolean status = false;
		System.out.print("\n Occurring elements of " + n + " times is : ");
		// Outer loop
		for (i = 0; i < size; ++i)
		{
			// Inner loop
			for (j = 0; j < size && counter >= 0 && counter <= n; ++j)
			{
				if (arr[i] == arr[j])
				{
					if (j < i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter++;
					}
				}
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				System.out.print("  " + arr[i]);
			}
			counter = 0;
		}
		if (status == false)
		{
			System.out.print("None");
		}
		System.out.print("\n\n");
	}
	public static void main(String[] args)
	{
		SearchElement s = new SearchElement();
		int[] arr = {
			2 , 7 , 1 , 8 , 3 , 0 , 4 , 1 , 4 , 3 , 6 , 2 , 5 , 6 , 1 , 2 , 3 , 6
		};
		// Get the size
		int size = arr.length;
		int n = 3;
		s.find_element(arr, n, size);
	}
}

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

// Include header file
#include <iostream>
using namespace std;
/*
  C++ Program
  Find the all elements that appears n times
*/
class SearchElement
{
	public:
		//Function which is display array elements
		void display(int arr[], int size)
		{
			for (int i = 0; i < size; ++i)
			{
				cout << "  " << arr[i];
			}
		}
	// Display the all elements which is occurs of n times in given collection
	void find_element(int arr[], int n, int size)
	{
		if (n <= 0)
		{
			return;
		}
		// Display element
		this->display(arr, size);
		// Loop controlling variables
		int i;
		int j;
		// Define counter variable
		int counter = 0;
		bool status = false;
		cout << "\n Occurring elements of " << n << " times is : ";
		// Outer loop
		for (i = 0; i < size; ++i)
		{
			// Inner loop
			for (j = 0; j < size && counter >= 0 && counter <= n; ++j)
			{
				if (arr[i] == arr[j])
				{
					if (j < i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter++;
					}
				}
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				cout << "  " << arr[i];
			}
			counter = 0;
		}
		if (status == false)
		{
			cout << "None";
		}
		cout << "\n\n";
	}
};
int main()
{
	SearchElement s = SearchElement();
	int arr[] = {
		2 , 7 , 1 , 8 , 3 , 0 , 4 , 1 , 4 , 3 , 6 , 2 , 5 , 6 , 1 , 2 , 3 , 6
	};
	// Get the size
	int size = sizeof(arr) / sizeof(arr[0]);
	int n = 3;
	s.find_element(arr, n, size);
	return 0;
}

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

// Include namespace system
using System;
/*
  C# Program
  Find the all elements that appears n times
*/
public class SearchElement
{
	//Function which is display array elements
	public void display(int[] arr, int size)
	{
		for (int i = 0; i < size; ++i)
		{
			Console.Write("  " + arr[i]);
		}
	}
	// Display the all elements which is occurs of n times in given collection
	public void find_element(int[] arr, int n, int size)
	{
		if (n <= 0)
		{
			return;
		}
		// Display element
		display(arr, size);
		// Loop controlling variables
		int i;
		int j;
		// Define counter variable
		int counter = 0;
		Boolean status = false;
		Console.Write("\n Occurring elements of " + n + " times is : ");
		// Outer loop
		for (i = 0; i < size; ++i)
		{
			// Inner loop
			for (j = 0; j < size && counter >= 0 && counter <= n; ++j)
			{
				if (arr[i] == arr[j])
				{
					if (j < i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter++;
					}
				}
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				Console.Write("  " + arr[i]);
			}
			counter = 0;
		}
		if (status == false)
		{
			Console.Write("None");
		}
		Console.Write("\n\n");
	}
	public static void Main(String[] args)
	{
		SearchElement s = new SearchElement();
		int[] arr = {
			2 , 7 , 1 , 8 , 3 , 0 , 4 , 1 , 4 , 3 , 6 , 2 , 5 , 6 , 1 , 2 , 3 , 6
		};
		// Get the size
		int size = arr.Length;
		int n = 3;
		s.find_element(arr, n, size);
	}
}

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

<?php
/*
  Php Program
  Find the all elements that appears n times
*/
class SearchElement
{
	//Function which is display array elements
	public	function display( & $arr, $size)
	{
		for ($i = 0; $i < $size; ++$i)
		{
			echo "  ". $arr[$i];
		}
	}
	// Display the all elements which is occurs of n times in given collection
	public	function find_element( & $arr, $n, $size)
	{
		if ($n <= 0)
		{
			return;
		}
		// Display element
		$this->display($arr, $size);
		// Loop controlling variables
		$i = 0;
		$j = 0;
		// Define counter variable
		$counter = 0;
		$status = false;
		echo "\n Occurring elements of ". $n ." times is : ";
		// Outer loop
		for ($i = 0; $i < $size; ++$i)
		{
			// Inner loop
			for ($j = 0; $j < $size && $counter >= 0 && $counter <= $n; ++$j)
			{
				if ($arr[$i] == $arr[$j])
				{
					if ($j < $i)
					{
						// If element has been already tested
						$counter = -1;
					}
					else
					{
						$counter++;
					}
				}
			}
			if ($counter == $n)
			{
				// When get new element
				$status = true;
				echo "  ". $arr[$i];
			}
			$counter = 0;
		}
		if ($status == false)
		{
			echo "None";
		}
		echo "\n\n";
	}
}

function main()
{
	$s = new SearchElement();
	$arr = array(2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6);
	// Get the size
	$size = count($arr);
	$n = 3;
	$s->find_element($arr, $n, $size);
}
main();

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

/*
  Node Js Program
  Find the all elements that appears n times
*/
class SearchElement
{
	//Function which is display array elements
	display(arr, size)
	{
		for (var i = 0; i < size; ++i)
		{
			process.stdout.write("  " + arr[i]);
		}
	}
	// Display the all elements which is occurs of n times in given collection
	find_element(arr, n, size)
	{
		if (n <= 0)
		{
			return;
		}
		// Display element
		this.display(arr, size);
		// Loop controlling variables
		var i = 0;
		var j = 0;
		// Define counter variable
		var counter = 0;
		var status = false;
		process.stdout.write("\n Occurring elements of " + n + " times is : ");
		// Outer loop
		for (i = 0; i < size; ++i)
		{
			// Inner loop
			for (j = 0; j < size && counter >= 0 && counter <= n; ++j)
			{
				if (arr[i] == arr[j])
				{
					if (j < i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter++;
					}
				}
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				process.stdout.write("  " + arr[i]);
			}
			counter = 0;
		}
		if (status == false)
		{
			process.stdout.write("None");
		}
		process.stdout.write("\n\n");
	}
}

function main()
{
	var s = new SearchElement();
	var arr = [2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6];
	// Get the size
	var size = arr.length;
	var n = 3;
	s.find_element(arr, n, size);
}
main();

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

#   Python 3 Program
#   Find the all elements that appears n times

class SearchElement :
	# Function which is display array elements
	def display(self, arr, size) :
		i = 0
		while (i < size) :
			print("  ", arr[i], end = "")
			i += 1
		
	
	#  Display the all elements which is occurs of n times in given collection
	def find_element(self, arr, n, size) :
		if (n <= 0) :
			return
		
		#  Display element
		self.display(arr, size)
		#  Loop controlling variables
		i = 0
		j = 0
		#  Define counter variable
		counter = 0
		status = False
		print("\n Occurring elements of ", n ," times is : ", end = "")
		#  Outer loop
		while (i < size) :
			#  Inner loop
			j = 0
			while (j < size and counter >= 0 and counter <= n) :
				if (arr[i] == arr[j]) :
					if (j < i) :
						#  If element has been already tested
						counter = -1
					else :
						counter += 1
					
				
				j += 1
			
			if (counter == n) :
				#  When get new element
				status = True
				print("  ", arr[i], end = "")
			
			counter = 0
			i += 1
		
		if (status == False) :
			print("None", end = "")
		
		print("\n")
	

def main() :
	s = SearchElement()
	arr = [2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6]
	#  Get the size
	size = len(arr)
	n = 3
	s.find_element(arr, n, size)

if __name__ == "__main__": main()

Output

   2   7   1   8   3   0   4   1   4   3   6   2   5   6   1   2   3   6
 Occurring elements of  3  times is :    2   1   3   6

#   Ruby Program
#   Find the all elements that appears n times

class SearchElement 
	# Function which is display array elements
	def display(arr, size) 
		i = 0
		while (i < size) 
			print("  ", arr[i])
			i += 1
		end

	end

	#  Display the all elements which is occurs of n times in given collection
	def find_element(arr, n, size) 
		if (n <= 0) 
			return
		end

		#  Display element
		self.display(arr, size)
		#  Loop controlling variables
		i = 0
		j = 0
		#  Define counter variable
		counter = 0
		status = false
		print("\n Occurring elements of ", n ," times is : ")
		#  Outer loop
		while (i < size) 
			#  Inner loop
			j = 0
			while (j < size && counter >= 0 && counter <= n) 
				if (arr[i] == arr[j]) 
					if (j < i) 
						#  If element has been already tested
						counter = -1
					else 
						counter += 1
					end

				end

				j += 1
			end

			if (counter == n) 
				#  When get new element
				status = true
				print("  ", arr[i])
			end

			counter = 0
			i += 1
		end

		if (status == false) 
			print("None")
		end

		print("\n\n")
	end

end

def main() 
	s = SearchElement.new()
	arr = [2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6]
	#  Get the size
	size = arr.length
	n = 3
	s.find_element(arr, n, size)
end

main()

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6


/*
  Scala Program
  Find the all elements that appears n times
*/
class SearchElement
{
	//Function which is display array elements
	def display(arr: Array[Int], size: Int): Unit = {
		var i: Int = 0;
		while (i < size)
		{
			print("  " + arr(i));
			i += 1;
		}
	}
	// Display the all elements which is occurs of n times in given collection
	def find_element(arr: Array[Int], n: Int, size: Int): Unit = {
		if (n <= 0)
		{
			return;
		}
		// Display element
		this.display(arr, size);
		// Loop controlling variables
		var i: Int = 0;
		var j: Int = 0;
		// Define counter variable
		var counter: Int = 0;
		var status: Boolean = false;
		print("\n Occurring elements of " + n + " times is : ");
		// Outer loop
		while (i < size)
		{
			// Inner loop
			j = 0;
			while (j < size && counter >= 0 && counter <= n)
			{
				if (arr(i) == arr(j))
				{
					if (j < i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter += 1;
					}
				}
				j += 1;
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				print("  " + arr(i));
			}
			counter = 0;
			i += 1;
		}
		if (status == false)
		{
			print("None");
		}
		print("\n\n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var s: SearchElement = new SearchElement();
		var arr: Array[Int] = Array(2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6);
		// Get the size
		var size: Int = arr.length;
		var n: Int = 3;
		s.find_element(arr, n, size);
	}
}

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6

/*
  Swift 4 Program
  Find the all elements that appears n times
*/
class SearchElement
{
	//Function which is display array elements
	func display(_ arr: [Int], _ size: Int)
	{
		var i: Int = 0;
		while (i < size)
		{
			print("  ", arr[i], terminator: "");
			i += 1;
		}
	}
	// Display the all elements which is occurs of n times in given collection
	func find_element(_ arr: [Int], _ n: Int, _ size: Int)
	{
		if (n <= 0)
		{
			return;
		}
		// Display element
		self.display(arr, size);
		// Loop controlling variables
		var i: Int = 0;
		var j: Int = 0;
		// Define counter variable
		var counter: Int = 0;
		var status: Bool = false;
		print("\n Occurring elements of ", n ," times is : ", terminator: "");
		// Outer loop
		while (i < size)
		{
			// Inner loop
			j = 0;
			while (j < size && counter >= 0 && counter <= n)
			{
				if (arr[i] == arr[j])
				{
					if (j < i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter += 1;
					}
				}
				j += 1;
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				print(" ", arr[i], terminator: "");
			}
			counter = 0;
			i += 1;
		}
		if (status == false)
		{
			print("None", terminator: "");
		}
		print("\n");
	}
}
func main()
{
	let s: SearchElement = SearchElement();
	let arr: [Int] = [2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6];
	// Get the size
	let size: Int = arr.count;
	let n: Int = 3;
	s.find_element(arr, n, size);
}
main();

Output

   2   7   1   8   3   0   4   1   4   3   6   2   5   6   1   2   3   6
 Occurring elements of  3  times is :   2  1  3  6

/*
  Kotlin Program
  Find the all elements that appears n times
*/
class SearchElement
{
	//Function which is display array elements
	fun display(arr: Array<Int>, size: Int): Unit
	{
		var i: Int = 0;
		while (i<size)
		{
			print("  " + arr[i]);
			i += 1;
		}
	}
	// Display the all elements which is occurs of n times in given collection
	fun find_element(arr: Array<Int>, n: Int, size: Int): Unit
	{
		if (n <= 0)
		{
			return;
		}
		// Display element
		this.display(arr, size);
		// Loop controlling variables
		var i: Int = 0;
		var j: Int ;
		// Define counter variable
		var counter: Int = 0;
		var status: Boolean = false;
		print("\n Occurring elements of " + n + " times is : ");
		// Outer loop
		while (i<size)
		{
			// Inner loop
			j = 0;
			while (j <size && counter>= 0 && counter <= n)
			{
				if (arr[i] == arr[j])
				{
					if (j<i)
					{
						// If element has been already tested
						counter = -1;
					}
					else
					{
						counter += 1;
					}
				}
				j += 1;
			}
			if (counter == n)
			{
				// When get new element
				status = true;
				print("  " + arr[i]);
			}
			counter = 0;
			i += 1;
		}
		if (status == false)
		{
			print("None");
		}
		print("\n\n");
	}
}
fun main(args: Array<String>): Unit
{
	var s: SearchElement = SearchElement();
	var arr: Array<Int> = arrayOf(2, 7, 1, 8, 3, 0, 4, 1, 4, 3, 6, 2, 5, 6, 1, 2, 3, 6);
	// Get the size
	var size: Int = arr.count();
	var n: Int = 3;
	s.find_element(arr, n, size);
}

Output

  2  7  1  8  3  0  4  1  4  3  6  2  5  6  1  2  3  6
 Occurring elements of 3 times is :   2  1  3  6
Last updated on by Kalkicode




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