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Check if permutation of number is power of 2 or not

Here given code implementation process.

// C program
// Check if permutation of number is power of 2 or not
#include <stdio.h>

// Check the both number have frequency is same or not
int isSameDigit(int num, int power)
{
	// Use to count number digit frequency
	// Digit [0 - 9]
	int a[10];
	int b[10];
	// Loop controlling variable
	int i = 0;
	// Set the initial frequency
	for (i = 0; i < 10; ++i)
	{
		a[i] = 0;
		b[i] = 0;
	}
	// Count digit frequency of num
	while (num > 0)
	{
		a[num % 10] += 1;
		// Remove last digit
		num = num / 10;
	}
	// Count digit frequency of power value
	while (power > 0)
	{
		b[power % 10]++;
		// Remove last digit
		power = power / 10;
	}
	// Compare frequency of both number
	for (i = 0; i < 10; ++i)
	{
		if (a[i] != b[i])
		{
			// When frequency are not same
			return 0;
		}
	}
	return 1;
}
void isPowerof2(int num)
{
	int result = 0;
	printf("\n Given num : %d", num);
	if (num > 0)
	{
		// Check all power of 2 in 32 bit integer
		for (int i = 0; i < 31 && result == 0; ++i)
		{
			if (num == (1 << i) || isSameDigit(num, 1 << i) == 1)
			{
				// Get resultant power
				result = (1 << i);
			}
		}
	}
	if (result != 0)
	{
		printf("\n Permutation %d is power of 2\n", result);
	}
	else
	{
		printf("\n No permutation of number %d is power of 2\n", num);
	}
}
int main(int argc, char
	const *argv[])
{
	// Test cases
	isPowerof2(76328);
	isPowerof2(1234);
	isPowerof2(8240);
	isPowerof2(23);
	isPowerof2(21);
	return 0;
}

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
// Java program for
// Check if permutation of number is power of 2 or not
public class PermutationPower
{
	// Check the both number have frequency is same or not
	public int isSameDigit(int num, int power)
	{
		// Use to count number digit frequency
		// Digit [0 - 9]
		int[] a = new int[10];
		int[] b = new int[10];
		// Loop controlling variable
		int i = 0;
		// Set the initial frequency
		for (i = 0; i < 10; ++i)
		{
			a[i] = 0;
			b[i] = 0;
		}
		// Count digit frequency of num
		while (num > 0)
		{
			a[num % 10] += 1;
			// Remove last digit
			num = num / 10;
		}
		// Count digit frequency of power value
		while (power > 0)
		{
			b[power % 10]++;
			// Remove last digit
			power = power / 10;
		}
		// Compare frequency of both number
		for (i = 0; i < 10; ++i)
		{
			if (a[i] != b[i])
			{
				// When frequency are not same
				return 0;
			}
		}
		return 1;
	}
	public void isPowerof2(int num)
	{
		int result = 0;
		System.out.print("\n Given num : " + num);
		if (num > 0)
		{
			// Check all power of 2 in 32 bit integer
			for (int i = 0; i < 31 && result == 0; ++i)
			{
				if (num == (1 << i) || isSameDigit(num, 1 << i) == 1)
				{
					// Get resultant power
					result = (1 << i);
				}
			}
		}
		if (result != 0)
		{
			System.out.print("\n Permutation " + result + " is power of 2\n");
		}
		else
		{
			System.out.print("\n No permutation of number " + num + " is power of 2\n");
		}
	}
	public static void main(String[] args)
	{
		PermutationPower task = new PermutationPower();
		// Test cases
		task.isPowerof2(76328);
		task.isPowerof2(1234);
		task.isPowerof2(8240);
		task.isPowerof2(23);
		task.isPowerof2(21);
	}
}

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
// Include header file
#include <iostream>
using namespace std;
// C++ program for
// Check if permutation of number is power of 2 or not

class PermutationPower
{
	public:
		// Check the both number have frequency is same or not
		int isSameDigit(int num, int power)
		{
			// Use to count number digit frequency
			// Digit [0 - 9]
			int a[10];
			int b[10];
			// Loop controlling variable
			int i = 0;
			// Set the initial frequency
			for (i = 0; i < 10; ++i)
			{
				a[i] = 0;
				b[i] = 0;
			}
			// Count digit frequency of num
			while (num > 0)
			{
				a[num % 10] += 1;
				// Remove last digit
				num = num / 10;
			}
			// Count digit frequency of power value
			while (power > 0)
			{
				b[power % 10]++;
				// Remove last digit
				power = power / 10;
			}
			// Compare frequency of both number
			for (i = 0; i < 10; ++i)
			{
				if (a[i] != b[i])
				{
					// When frequency are not same
					return 0;
				}
			}
			return 1;
		}
	void isPowerof2(int num)
	{
		int result = 0;
		cout << "\n Given num : " << num;
		if (num > 0)
		{
			// Check all power of 2 in 32 bit integer
			for (int i = 0; i < 31 && result == 0; ++i)
			{
				if (num == (1 << i) || this->isSameDigit(num, 1 << i) == 1)
				{
					// Get resultant power
					result = (1 << i);
				}
			}
		}
		if (result != 0)
		{
			cout << "\n Permutation " << result << " is power of 2\n";
		}
		else
		{
			cout << "\n No permutation of number " << num << " is power of 2\n";
		}
	}
};
int main()
{
	PermutationPower *task = new PermutationPower();
	// Test cases
	task->isPowerof2(76328);
	task->isPowerof2(1234);
	task->isPowerof2(8240);
	task->isPowerof2(23);
	task->isPowerof2(21);
	return 0;
}

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
// Include namespace system
using System;
// Csharp program for
// Check if permutation of number is power of 2 or not
public class PermutationPower
{
	// Check the both number have frequency is same or not
	public int isSameDigit(int num, int power)
	{
		// Use to count number digit frequency
		// Digit [0 - 9]
		int[] a = new int[10];
		int[] b = new int[10];
		// Loop controlling variable
		int i = 0;
		// Set the initial frequency
		for (i = 0; i < 10; ++i)
		{
			a[i] = 0;
			b[i] = 0;
		}
		// Count digit frequency of num
		while (num > 0)
		{
			a[num % 10] += 1;
			// Remove last digit
			num = num / 10;
		}
		// Count digit frequency of power value
		while (power > 0)
		{
			b[power % 10]++;
			// Remove last digit
			power = power / 10;
		}
		// Compare frequency of both number
		for (i = 0; i < 10; ++i)
		{
			if (a[i] != b[i])
			{
				// When frequency are not same
				return 0;
			}
		}
		return 1;
	}
	public void isPowerof2(int num)
	{
		int result = 0;
		Console.Write("\n Given num : " + num);
		if (num > 0)
		{
			// Check all power of 2 in 32 bit integer
			for (int i = 0; i < 31 && result == 0; ++i)
			{
				if (num == (1 << i) || this.isSameDigit(num, 1 << i) == 1)
				{
					// Get resultant power
					result = (1 << i);
				}
			}
		}
		if (result != 0)
		{
			Console.Write("\n Permutation " + result + " is power of 2\n");
		}
		else
		{
			Console.Write("\n No permutation of number " + num + " is power of 2\n");
		}
	}
	public static void Main(String[] args)
	{
		PermutationPower task = new PermutationPower();
		// Test cases
		task.isPowerof2(76328);
		task.isPowerof2(1234);
		task.isPowerof2(8240);
		task.isPowerof2(23);
		task.isPowerof2(21);
	}
}

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
<?php
// Php program for
// Check if permutation of number is power of 2 or not
class PermutationPower
{
	// Check the both number have frequency is same or not
	public	function isSameDigit($num, $power)
	{
		// Use to count number digit frequency
		// Digit [0 - 9]
		// Set the initial frequency
		$a = array_fill(0, 10, 0);
		$b = array_fill(0, 10, 0);
		// Loop controlling variable
		$i = 0;
		// Count digit frequency of num
		while ($num > 0)
		{
			$a[$num % 10] += 1;
			// Remove last digit
			$num = (int)($num / 10);
		}
		// Count digit frequency of power value
		while ($power > 0)
		{
			$b[$power % 10]++;
			// Remove last digit
			$power = (int)($power / 10);
		}
		// Compare frequency of both number
		for ($i = 0; $i < 10; ++$i)
		{
			if ($a[$i] != $b[$i])
			{
				// When frequency are not same
				return 0;
			}
		}
		return 1;
	}
	public	function isPowerof2($num)
	{
		$result = 0;
		echo("\n Given num : ".$num);
		if ($num > 0)
		{
			// Check all power of 2 in 32 bit integer
			for ($i = 0; $i < 31 && $result == 0; ++$i)
			{
				if ($num == (1 << $i) || $this->isSameDigit($num, 1 << $i) == 1)
				{
					// Get resultant power
					$result = (1 << $i);
				}
			}
		}
		if ($result != 0)
		{
			echo("\n Permutation ".$result.
				" is power of 2\n");
		}
		else
		{
			echo("\n No permutation of number ".$num.
				" is power of 2\n");
		}
	}
}

function main()
{
	$task = new PermutationPower();
	// Test cases
	$task->isPowerof2(76328);
	$task->isPowerof2(1234);
	$task->isPowerof2(8240);
	$task->isPowerof2(23);
	$task->isPowerof2(21);
}
main();

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
// Node JS program for
// Check if permutation of number is power of 2 or not
class PermutationPower
{
	// Check the both number have frequency is same or not
	isSameDigit(num, power)
	{
		// Use to count number digit frequency
		// Digit [0 - 9]
		// Set the initial frequency
		var a = Array(10).fill(0);
		var b = Array(10).fill(0);
		// Loop controlling variable
		var i = 0;
		// Count digit frequency of num
		while (num > 0)
		{
			a[num % 10] += 1;
			// Remove last digit
			num = parseInt(num / 10);
		}
		// Count digit frequency of power value
		while (power > 0)
		{
			b[power % 10]++;
			// Remove last digit
			power = parseInt(power / 10);
		}
		// Compare frequency of both number
		for (i = 0; i < 10; ++i)
		{
			if (a[i] != b[i])
			{
				// When frequency are not same
				return 0;
			}
		}
		return 1;
	}
	isPowerof2(num)
	{
		var result = 0;
		process.stdout.write("\n Given num : " + num);
		if (num > 0)
		{
			// Check all power of 2 in 32 bit integer
			for (var i = 0; i < 31 && result == 0; ++i)
			{
				if (num == (1 << i) || this.isSameDigit(num, 1 << i) == 1)
				{
					// Get resultant power
					result = (1 << i);
				}
			}
		}
		if (result != 0)
		{
			process.stdout.write("\n Permutation " + result + " is power of 2\n");
		}
		else
		{
			process.stdout.write("\n No permutation of number " + num + " is power of 2\n");
		}
	}
}

function main()
{
	var task = new PermutationPower();
	// Test cases
	task.isPowerof2(76328);
	task.isPowerof2(1234);
	task.isPowerof2(8240);
	task.isPowerof2(23);
	task.isPowerof2(21);
}
main();

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
#  Python 3 program for
#  Check if permutation of number is power of 2 or not
class PermutationPower :
	#  Check the both number have frequency is same or not
	def isSameDigit(self, num, power) :
		#  Use to count number digit frequency
		#  Digit [0 - 9]
		#  Set the initial frequency
		a = [0] * (10)
		b = [0] * (10)
		#  Loop controlling variable
		i = 0
		#  Count digit frequency of num
		while (num > 0) :
			a[num % 10] += 1
			#  Remove last digit
			num = int(num / 10)
		
		#  Count digit frequency of power value
		while (power > 0) :
			b[power % 10] += 1
			#  Remove last digit
			power = int(power / 10)
		
		#  Compare frequency of both number
		i = 0
		while (i < 10) :
			if (a[i] != b[i]) :
				#  When frequency are not same
				return 0
			
			i += 1
		
		return 1
	
	def isPowerof2(self, num) :
		result = 0
		print("\n Given num : ", num, end = "")
		if (num > 0) :
			#  Check all power of 2 in 32 bit integer
			i = 0
			while (i < 31 and result == 0) :
				if (num == (1 << i) or self.isSameDigit(num, 1 << i) == 1) :
					#  Get resultant power
					result = (1 << i)
				
				i += 1
			
		
		if (result != 0) :
			print("\n Permutation", result ,"is power of 2")
		else :
			print("\n No permutation of number", num ,"is power of 2")
		
	

def main() :
	task = PermutationPower()
	#  Test cases
	task.isPowerof2(76328)
	task.isPowerof2(1234)
	task.isPowerof2(8240)
	task.isPowerof2(23)
	task.isPowerof2(21)

if __name__ == "__main__": main()

input

 Given num :  76328
 Permutation 32768 is power of 2

 Given num :  1234
 No permutation of number 1234 is power of 2

 Given num :  8240
 Permutation 2048 is power of 2

 Given num :  23
 Permutation 32 is power of 2

 Given num :  21
 No permutation of number 21 is power of 2
#  Ruby program for
#  Check if permutation of number is power of 2 or not
class PermutationPower 
	#  Check the both number have frequency is same or not
	def isSameDigit(num, power) 
		#  Use to count number digit frequency
		#  Digit [0 - 9]
		#  Set the initial frequency
		a = Array.new(10) {0}
		b = Array.new(10) {0}
		#  Loop controlling variable
		i = 0
		#  Count digit frequency of num
		while (num > 0) 
			a[num % 10] += 1
			#  Remove last digit
			num = num / 10
		end

		#  Count digit frequency of power value
		while (power > 0) 
			b[power % 10] += 1
			#  Remove last digit
			power = power / 10
		end

		#  Compare frequency of both number
		i = 0
		while (i < 10) 
			if (a[i] != b[i]) 
				#  When frequency are not same
				return 0
			end

			i += 1
		end

		return 1
	end

	def isPowerof2(num) 
		result = 0
		print("\n Given num : ", num)
		if (num > 0) 
			#  Check all power of 2 in 32 bit integer
			i = 0
			while (i < 31 && result == 0) 
				if (num == (1 << i) || self.isSameDigit(num, 1 << i) == 1) 
					#  Get resultant power
					result = (1 << i)
				end

				i += 1
			end

		end

		if (result != 0) 
			print("\n Permutation ", result ," is power of 2\n")
		else 
			print("\n No permutation of number ", num ," is power of 2\n")
		end

	end

end

def main() 
	task = PermutationPower.new()
	#  Test cases
	task.isPowerof2(76328)
	task.isPowerof2(1234)
	task.isPowerof2(8240)
	task.isPowerof2(23)
	task.isPowerof2(21)
end

main()

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2

// Scala program for
// Check if permutation of number is power of 2 or not
class PermutationPower()
{
	// Check the both number have frequency is same or not
	def isSameDigit(n: Int, p: Int): Int = {
      	var num = n;
      	var power = p;
		// Use to count number digit frequency
		// Digit [0 - 9]
		// Set the initial frequency
		var a: Array[Int] = Array.fill[Int](10)(0);
		var b: Array[Int] = Array.fill[Int](10)(0);
		// Loop controlling variable
		var i: Int = 0;
		// Count digit frequency of num
		while (num > 0)
		{
			a(num % 10) += 1;
			// Remove last digit
			num = num / 10;
		}
		// Count digit frequency of power value
		while (power > 0)
		{
			b(power % 10) += 1;
			// Remove last digit
			power = power / 10;
		}
		// Compare frequency of both number
		while (i < 10)
		{
			if (a(i) != b(i))
			{
				// When frequency are not same
				return 0;
			}
			i += 1;
		}
		return 1;
	}
	def isPowerof2(num: Int): Unit = {
		var result: Int = 0;
		print("\n Given num : " + num);
		if (num > 0)
		{
			// Check all power of 2 in 32 bit integer
			var i: Int = 0;
			while (i < 31 && result == 0)
			{
				if (num == (1 << i) || isSameDigit(num, 1 << i) == 1)
				{
					// Get resultant power
					result = (1 << i);
				}
				i += 1;
			}
		}
		if (result != 0)
		{
			print("\n Permutation " + result + " is power of 2\n");
		}
		else
		{
			print("\n No permutation of number " + num + " is power of 2\n");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: PermutationPower = new PermutationPower();
		// Test cases
		task.isPowerof2(76328);
		task.isPowerof2(1234);
		task.isPowerof2(8240);
		task.isPowerof2(23);
		task.isPowerof2(21);
	}
}

input

 Given num : 76328
 Permutation 32768 is power of 2

 Given num : 1234
 No permutation of number 1234 is power of 2

 Given num : 8240
 Permutation 2048 is power of 2

 Given num : 23
 Permutation 32 is power of 2

 Given num : 21
 No permutation of number 21 is power of 2
// Swift 4 program for
// Check if permutation of number is power of 2 or not
class PermutationPower
{
	// Check the both number have frequency is same or not
	func isSameDigit(_ n: Int, _ p: Int) -> Int
	{
      	var num = n;
      	var power = p;
		// Use to count number digit frequency
		// Digit [0 - 9]
		// Set the initial frequency
		var a: [Int] = Array(repeating: 0, count: 10);
		var b: [Int] = Array(repeating: 0, count: 10);
		// Loop controlling variable
		var i: Int = 0;
		// Count digit frequency of num
		while (num > 0)
		{
			a[num % 10] += 1;
			// Remove last digit
			num = num / 10;
		}
		// Count digit frequency of power value
		while (power > 0)
		{
			b[power % 10] += 1;
			// Remove last digit
			power = power / 10;
		}
		// Compare frequency of both number
		i = 0;
		while (i < 10)
		{
			if (a[i] != b[i])
			{
				// When frequency are not same
				return 0;
			}
			i += 1;
		}
		return 1;
	}
	func isPowerof2(_ num: Int)
	{
		var result: Int = 0;
		print("\n Given num : ", num, terminator: "");
		if (num > 0)
		{
			// Check all power of 2 in 32 bit integer
			var i: Int = 0;
			while (i < 31 && result == 0)
			{
				if (num == (1 << i) || self.isSameDigit(num, 1 << i) == 1)
				{
					// Get resultant power
					result = (1 << i);
				}
				i += 1;
			}
		}
		if (result != 0)
		{
			print("\n Permutation ", result ," is power of 2");
		}
		else
		{
			print("\n No permutation of number ", num ," is power of 2");
		}
	}
}
func main()
{
	let task: PermutationPower = PermutationPower();
	// Test cases
	task.isPowerof2(76328);
	task.isPowerof2(1234);
	task.isPowerof2(8240);
	task.isPowerof2(23);
	task.isPowerof2(21);
}
main();

input

 Given num :  76328
 Permutation  32768  is power of 2

 Given num :  1234
 No permutation of number  1234  is power of 2

 Given num :  8240
 Permutation  2048  is power of 2

 Given num :  23
 Permutation  32  is power of 2

 Given num :  21
 No permutation of number  21  is power of 2
Last updated on by Kalkicode




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