Find the element that appears once in array Set B

Given of collection of integer elements. That is contains one single element and remaining elements are appears exactly three three times. Our goal is to find that single element which is single of this collection.

Here given code implementation process.

// C Program
// Find the element that appears once in array 
// When other elements are exactly repeating of 3 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]);
	}
}
// Finds the single appearing element which is exist in repeating element
void single_element(int arr[], int size)
{
	int i = 0;
	// Define useful resultant variables
	// Get first element
	int ones = 0;
	int twos = 0;
	int threes = 0;
	//Execute Loop is from 0 ... size
	for (i = 0; i < size; ++i)
	{
		twos = twos | (ones & arr[i]);
		ones = ones ^ arr[i];
		threes = (ones & twos);
		ones = ones & (~threes);
		twos = twos & (~threes);
	}
	display(arr, size);
	// Display calculated result
	printf("\n  Single element = %d \n", ones);
}
int main()
{
	// Define Array which is containing one element occurrences single and 
	// Other elements are exist 3-3 times
	int arr1[] = {
		1 , 2 , 3 , 8 , 1 , 4 , 2 , 4 , 1 , 3 , 2 , 3 , 4
	};
	int arr2[] = {
		1 , 2 , 2 , 2 , 3 , 1 , 1 , 4 , 4 , 4
	};
	// Get the size
	int size = sizeof(arr1) / sizeof(arr1[0]);
	single_element(arr1, size);
	// Get the size
	size = sizeof(arr2) / sizeof(arr2[0]);
	single_element(arr2, size);
	return 0;
}

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
  Single element = 8
  1  2  2  2  3  1  1  4  4  4
  Single element = 3
/*
  Java Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
public class SingleElement
{
	//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]);
		}
	}
	// Finds the single appearing element which is exist in repeating element
	public void single_element(int[] arr, int size)
	{
		int i = 0;
		// Define useful resultant variables
		// Get first element
		int ones = 0;
		int twos = 0;
		int threes = 0;
		//Execute Loop is from 0... size
		for (i = 0; i < size; ++i)
		{
			twos = twos | (ones & arr[i]);
			ones = ones ^ arr[i];
			threes = (ones & twos);
			ones = ones & (~threes);
			twos = twos & (~threes);
		}
		display(arr, size);
		// Display calculated result
		System.out.print("\n Single element = " + ones + " \n");
	}
	public static void main(String[] args)
	{
		SingleElement s = new SingleElement();
		// Define Array which is containing one element occurrences single and 
		// Other elements are exist 3-3 times
		int[] arr1 = {
			1 , 2 , 3 , 8 , 1 , 4 , 2 , 4 , 1 , 3 , 2 , 3 , 4
		};
		int[] arr2 = {
			1 , 2 , 2 , 2 , 3 , 1 , 1 , 4 , 4 , 4
		};
		// Get the size
		int size = arr1.length;
		s.single_element(arr1, size);
		// Get the size
		size = arr2.length;
		s.single_element(arr2, size);
	}
}

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3
// Include header file
#include <iostream>

using namespace std;
/*
  C++ Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
class SingleElement
{
	public:
		//Function which is display array elements
		void display(int arr[], int size)
		{
			for (int i = 0; i < size; ++i)
			{
				cout << "  " << arr[i];
			}
		}
	// Finds the single appearing element which is exist in repeating element
	void single_element(int arr[], int size)
	{
		int i = 0;
		// Define useful resultant variables
		// Get first element
		int ones = 0;
		int twos = 0;
		int threes = 0;
		//Execute Loop is from 0... size
		for (i = 0; i < size; ++i)
		{
			twos = twos | (ones & arr[i]);
			ones = ones ^ arr[i];
			threes = (ones & twos);
			ones = ones & (~threes);
			twos = twos & (~threes);
		}
		this->display(arr, size);
		// Display calculated result
		cout << "\n Single element = " << ones << " \n";
	}
};
int main()
{
	SingleElement s = SingleElement();
	// Define Array which is containing one element occurrences single and
	// Other elements are exist 3-3 times
	int arr1[] = {
		1 , 2 , 3 , 8 , 1 , 4 , 2 , 4 , 1 , 3 , 2 , 3 , 4
	};
	int arr2[] = {
		1 , 2 , 2 , 2 , 3 , 1 , 1 , 4 , 4 , 4
	};
	// Get the size
	int size = sizeof(arr1) / sizeof(arr1[0]);
	s.single_element(arr1, size);
	// Get the size
	size = sizeof(arr2) / sizeof(arr2[0]);
	s.single_element(arr2, size);
	return 0;
}

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3
// Include namespace system
using System;
/*
  C# Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
public class SingleElement
{
	//Function which is display array elements
	public void display(int[] arr, int size)
	{
		for (int i = 0; i < size; ++i)
		{
			Console.Write("  " + arr[i]);
		}
	}
	// Finds the single appearing element which is exist in repeating element
	public void single_element(int[] arr, int size)
	{
		int i = 0;
		// Define useful resultant variables
		// Get first element
		int ones = 0;
		int twos = 0;
		int threes = 0;
		//Execute Loop is from 0... size
		for (i = 0; i < size; ++i)
		{
			twos = twos | (ones & arr[i]);
			ones = ones ^ arr[i];
			threes = (ones & twos);
			ones = ones & (~threes);
			twos = twos & (~threes);
		}
		display(arr, size);
		// Display calculated result
		Console.Write("\n Single element = " + ones + " \n");
	}
	public static void Main(String[] args)
	{
		SingleElement s = new SingleElement();
		// Define Array which is containing one element occurrences single and
		// Other elements are exist 3-3 times
		int[] arr1 = {
			1 , 2 , 3 , 8 , 1 , 4 , 2 , 4 , 1 , 3 , 2 , 3 , 4
		};
		int[] arr2 = {
			1 , 2 , 2 , 2 , 3 , 1 , 1 , 4 , 4 , 4
		};
		// Get the size
		int size = arr1.Length;
		s.single_element(arr1, size);
		// Get the size
		size = arr2.Length;
		s.single_element(arr2, size);
	}
}

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3
<?php
/*
  Php Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
class SingleElement
{
	//Function which is display array elements
	public	function display( & $arr, $size)
	{
		for ($i = 0; $i < $size; ++$i)
		{
			echo "  ". $arr[$i];
		}
	}
	// Finds the single appearing element which is exist in repeating element
	public	function single_element( & $arr, $size)
	{
		$i = 0;
		// Define useful resultant variables
		// Get first element
		$ones = 0;
		$twos = 0;
		$threes = 0;
		//Execute Loop is from 0... size
		for ($i = 0; $i < $size; ++$i)
		{
			$twos = $twos | ($ones & $arr[$i]);
			$ones = $ones ^ $arr[$i];
			$threes = ($ones & $twos);
			$ones = $ones & (~$threes);
			$twos = $twos & (~$threes);
		}
		$this->display($arr, $size);
		// Display calculated result
		echo "\n Single element = ". $ones ." \n";
	}
}

function main()
{
	$s = new SingleElement();
	// Define Array which is containing one element occurrences single and
	// Other elements are exist 3-3 times
	$arr1 = array(1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4);
	$arr2 = array(1, 2, 2, 2, 3, 1, 1, 4, 4, 4);
	// Get the size
	$size = count($arr1);
	$s->single_element($arr1, $size);
	// Get the size
	$size = count($arr2);
	$s->single_element($arr2, $size);
}
main();

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3
/*
  Node Js Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
class SingleElement
{
	//Function which is display array elements
	display(arr, size)
	{
		for (var i = 0; i < size; ++i)
		{
			process.stdout.write("  " + arr[i]);
		}
	}
	// Finds the single appearing element which is exist in repeating element
	single_element(arr, size)
	{
		var i = 0;
		// Define useful resultant variables
		// Get first element
		var ones = 0;
		var twos = 0;
		var threes = 0;
		//Execute Loop is from 0... size
		for (i = 0; i < size; ++i)
		{
			twos = twos | (ones & arr[i]);
			ones = ones ^ arr[i];
			threes = (ones & twos);
			ones = ones & (~threes);
			twos = twos & (~threes);
		}
		this.display(arr, size);
		// Display calculated result
		process.stdout.write("\n Single element = " + ones + " \n");
	}
}

function main()
{
	var s = new SingleElement();
	// Define Array which is containing one element occurrences single and
	// Other elements are exist 3-3 times
	var arr1 = [1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4];
	var arr2 = [1, 2, 2, 2, 3, 1, 1, 4, 4, 4];
	// Get the size
	var size = arr1.length;
	s.single_element(arr1, size);
	// Get the size
	size = arr2.length;
	s.single_element(arr2, size);
}
main();

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3
#   Python 3 Program
#   Find the element that appears once in array
#   When other elements are exactly repeating of 3 times

class SingleElement :
	# Function which is display array elements
	def display(self, arr, size) :
		i = 0
		while (i < size) :
			print("  ", arr[i], end = "")
			i += 1
		
	
	#  Finds the single appearing element which is exist in repeating element
	def single_element(self, arr, size) :
		i = 0
		#  Define useful resultant variables
		#  Get first element
		ones = 0
		twos = 0
		threes = 0
		# Execute Loop is from 0... size
		while (i < size) :
			twos = twos | (ones & arr[i])
			ones = ones ^ arr[i]
			threes = (ones & twos)
			ones = ones & (~threes)
			twos = twos & (~threes)
			i += 1
		
		self.display(arr, size)
		#  Display calculated result
		print("\n Single element = ", ones ," ")
	

def main() :
	s = SingleElement()
	#  Define Array which is containing one element occurrences single and
	#  Other elements are exist 3-3 times
	arr1 = [1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4]
	arr2 = [1, 2, 2, 2, 3, 1, 1, 4, 4, 4]
	#  Get the size
	size = len(arr1)
	s.single_element(arr1, size)
	#  Get the size
	size = len(arr2)
	s.single_element(arr2, size)

if __name__ == "__main__": main()

Output

   1   2   3   8   1   4   2   4   1   3   2   3   4
 Single element =  8
   1   2   2   2   3   1   1   4   4   4
 Single element =  3
#   Ruby Program
#   Find the element that appears once in array
#   When other elements are exactly repeating of 3 times

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

	end

	#  Finds the single appearing element which is exist in repeating element
	def single_element(arr, size) 
		i = 0
		#  Define useful resultant variables
		#  Get first element
		ones = 0
		twos = 0
		threes = 0
		# Execute Loop is from 0... size
		while (i < size) 
			twos = twos | (ones & arr[i])
			ones = ones ^ arr[i]
			threes = (ones & twos)
			ones = ones & (~threes)
			twos = twos & (~threes)
			i += 1
		end

		self.display(arr, size)
		#  Display calculated result
		print("\n Single element = ", ones ," \n")
	end

end

def main() 
	s = SingleElement.new()
	#  Define Array which is containing one element occurrences single and
	#  Other elements are exist 3-3 times
	arr1 = [1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4]
	arr2 = [1, 2, 2, 2, 3, 1, 1, 4, 4, 4]
	#  Get the size
	size = arr1.length
	s.single_element(arr1, size)
	#  Get the size
	size = arr2.length
	s.single_element(arr2, size)
end

main()

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8 
  1  2  2  2  3  1  1  4  4  4
 Single element = 3 
/*
  Scala Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
class SingleElement
{
	//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;
		}
	}
	// Finds the single appearing element which is exist in repeating element
	def single_element(arr: Array[Int], size: Int): Unit = {
		var i: Int = 0;
		// Define useful resultant variables
		// Get first element
		var ones: Int = 0;
		var twos: Int = 0;
		var threes: Int = 0;
		//Execute Loop is from 0... size
		while (i < size)
		{
			twos = twos | (ones & arr(i));
			ones = ones ^ arr(i);
			threes = (ones & twos);
			ones = ones & (~threes);
			twos = twos & (~threes);
			i += 1;
		}
		this.display(arr, size);
		// Display calculated result
		print("\n Single element = " + ones + " \n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var s: SingleElement = new SingleElement();
		// Define Array which is containing one element occurrences single and
		// Other elements are exist 3-3 times
		var arr1: Array[Int] = Array(1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4);
		var arr2: Array[Int] = Array(1, 2, 2, 2, 3, 1, 1, 4, 4, 4);
		// Get the size
		var size: Int = arr1.length;
		s.single_element(arr1, size);
		// Get the size
		size = arr2.length;
		s.single_element(arr2, size);
	}
}

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3
/*
  Swift 4 Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
class SingleElement
{
	//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;
		}
	}
	// Finds the single appearing element which is exist in repeating element
	func single_element(_ arr: [Int], _ size: Int)
	{
		var i: Int = 0;
		// Define useful resultant variables
		// Get first element
		var ones: Int = 0;
		var twos: Int = 0;
		var threes: Int = 0;
		//Execute Loop is from 0... size
		while (i < size)
		{
			twos = twos | (ones & arr[i]);
			ones = ones ^ arr[i];
			threes = (ones & twos);
			ones = ones & (~threes);
			twos = twos & (~threes);
			i += 1;
		}
		self.display(arr, size);
		// Display calculated result
		print("\n Single element = ", ones ," ");
	}
}
func main()
{
	let s: SingleElement = SingleElement();
	// Define Array which is containing one element occurrences single and
	// Other elements are exist 3-3 times
	let arr1: [Int] = [1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4];
	let arr2: [Int] = [1, 2, 2, 2, 3, 1, 1, 4, 4, 4];
	// Get the size
	var size: Int = arr1.count;
	s.single_element(arr1, size);
	// Get the size
	size = arr2.count;
	s.single_element(arr2, size);
}
main();

Output

   1   2   3   8   1   4   2   4   1   3   2   3   4
 Single element =  8
   1   2   2   2   3   1   1   4   4   4
 Single element =  3
/*
  Kotlin Program
  Find the element that appears once in array
  When other elements are exactly repeating of 3 times
*/
class SingleElement
{
	//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;
		}
	}
	// Finds the single appearing element which is exist in repeating element
	fun single_element(arr: Array<Int>, size: Int): Unit
	{
		var i: Int = 0;
		// Define useful resultant variables
		// Get first element
		var ones: Int = 0;
		var twos: Int = 0;
		var threes: Int;
		//Execute Loop is from 0... size
		while (i<size)
		{
			twos = twos or(ones and arr[i]);
			ones = ones xor arr[i];
			threes = (ones and twos);
			ones = ones and threes.inv();
			twos = twos and threes.inv();
			i += 1;
		}
		this.display(arr, size);
		// Display calculated result
		print("\n Single element = " + ones + " \n");
	}
}
fun main(args: Array<String>): Unit
{
	var s: SingleElement = SingleElement();
	// Define Array which is containing one element occurrences single and
	// Other elements are exist 3-3 times
	var arr1: Array<Int> = arrayOf(1, 2, 3, 8, 1, 4, 2, 4, 1, 3, 2, 3, 4);
	var arr2: Array<Int> = arrayOf(1, 2, 2, 2, 3, 1, 1, 4, 4, 4);
	// Get the size
	var size: Int = arr1.count();
	s.single_element(arr1, size);
	// Get the size
	size = arr2.count();
	s.single_element(arr2, size);
}

Output

  1  2  3  8  1  4  2  4  1  3  2  3  4
 Single element = 8
  1  2  2  2  3  1  1  4  4  4
 Single element = 3


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