Given a number check whether it is a power of 8 or not.
Examples :
Input : n = 64 Output : Yes Input : 75 Output : No
First solution
We calculate log8(n) of the number if it is an integer, then n is in the power of 8. We use trunc(n) function that finds the closest integer for a double value.
C++
// C++ program to check if a number is power of 8 #include <cmath> #include <iostream> using namespace std; /* function to check if power of 8 */ bool checkPowerof8( int n) { /* calculate log8(n) */ double i = log (n) / log (8); /* check if i is an integer or not */ return (i - trunc(i) < 0.000001); } /* driver function */ int main() { int n = 65; checkPowerof8(n) ? cout << "Yes" : cout << "No" ; return 0; } |
Java
// Java program to check if // a number is power of 8 class GFG { // function to check // if power of 8 static boolean checkPowerof8( int n) { /* calculate log8(n) */ double i = Math.log(n) / Math.log( 8 ); /* check if i is an integer or not */ return (i - Math.floor(i) < 0.000001 ); } // Driver Code public static void main(String args[]) { int n = 65 ; if (checkPowerof8(n)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by Sam007 |
Python3
# Python3 program to check # if a number is power of 8 from math import log,trunc # function to check if power of 8 def checkPowerof8(n): # calculate log8(n) i = log(n, 8 ) # check if i is an integer or not return (i - trunc(i) < 0.000001 ); # Driver Code n = 65 if checkPowerof8(n): print ( "Yes" ) else : print ( "No" ) # This code is contributed by Mohit Kumar |
C#
// C# program to check if // a number is power of 8 using System; class GFG { // function to check // if power of 8 static bool checkPowerof8( int n) { // calculate log8(n) */ double i = Math.Log(n) / Math.Log(8); // check if i is an integer or not */ return (i - Math.Floor(i) < 0.000001); } // Driver Code static public void Main() { int n = 65; if (checkPowerof8(n)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by akt_mit |
PHP
<?php // PHP program to check if // a number is power of 8 // function to check // if power of 8 function checkPowerof8( $n ) { /* calculate log8(n) */ $i = log( $n ) / log(8); /* check if i is an integer or not */ return ( $i - floor ( $i ) < 0.000001); } // Driver Code $n = 65; if (checkPowerof8( $n )) echo "Yes" ; else echo "No" ; // This code is contributed // by Sach_Code ?> |
Javascript
<script> // Javascript program to check if // a number is power of 8 // function to check // if power of 8 function checkPowerof8(n) { // calculate log8(n) */ let i = Math.log(n) / Math.log(8); // check if i is an integer or not */ return (i - Math.floor(i) < 0.000001); } let n = 65; if (checkPowerof8(n)) document.write( "Yes" ); else document.write( "No" ); </script> |
Output :
No
Time Complexity: O(1)
Auxiliary Space: O(1)
Second solution
A number is a power of 8 if the following conditions are satisfied.
- The number is the power of two. A number is the power of two if it has only one set bit, i.e., bitwise and of n and n-1 is 0.
- The number has its only set a bit at position 0 or 3 or 6 or …. 30 [For a 32-bit number]. To check the position of its set bit we can use a mask (0xB6DB6DB6)16 = (10110110110110110110110110110110)2.
Below is the implementation of the above idea.
C++
// C++ program to check if a number is power of 8 // using bit mask. #include <bits/stdc++.h> using namespace std; /*function to check if power of 8*/ bool checkPowerof8( int n) { return (n && !(n & (n - 1)) && !(n & 0xB6DB6DB6)); } /*driver function*/ int main() { int n = 65; checkPowerof8(n) ? cout << "Yes" : cout << "No" ; return 0; } |
Java
// Java program to check if a // number is power of 8 using // bit mask. import java.util.*; class GFG{ // function to check if // power of 8 static boolean checkPowerof8( int n) { return (n > 0 && (n & (n - 1 )) > 0 && (n & 0xB6DB6DB6 ) > 0 ); } // Driver code public static void main(String[] args) { int n = 65 ; if (checkPowerof8(n) == true ) System.out.print( "Yes" ); else System.out.print( "No" ); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to check if a number # is power of 8 # function to check if power of 8 def checkPowerof8(n): return (n and not (n & (n - 1 )) and not (n & 0xB6DB6DB6 )) # Driver Code if __name__ = = "__main__" : n = 65 if checkPowerof8(n): print ( "Yes" ) else : print ( "No" ) # This code is contributed by ita_c |
C#
// C# program to check if a // number is power of 8 using // bit mask. using System; class GFG{ // Function to check if // power of 8 static bool checkPowerof8( int n) { return (n > 0 && (n & (n - 1)) > 0 && (n & 0xB6DB6DB6) > 0); } // Driver code static public void Main() { int n = 65; if (checkPowerof8(n) == true ) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } } // This code is contributed by avanitrachhadiya2155 |
PHP
<?php // PHP program to check if a number // is power of 8 using bit mask. // function to check if power of 8 function checkPowerof8( $n ) { $t = ( $n && !( $n & ( $n - 1)) && !( $n & 0xB6DB6DB6)); return $t ; } // Driver Code $n = 65; if (checkPowerof8( $n )) echo "Yes" ; else echo "No" ; // This code is contributed by Sach ?> |
Javascript
<script> /*function to check if power of 8*/ function checkPowerof8( n) { return (n && !(n & (n - 1)) && !(n & 0xB6DB6DB6)); } var n = 65; if (checkPowerof8(n)) document.write( "Yes" ); else document.write( "No" ); </script> |
Output :
No
Time Complexity: O(1)
Auxiliary Space: O(1)
One simple observation that can be made here is that if a number is the power of 8 then it has only a one-bit set and that bit is at positions 1, 4, 7, 10, …
Thus, we can just check if the only bit set in the number is at one of these positions then it is a power of 8 otherwise not.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to check if n is power of 8 bool checkPowerof8( int n) { // Variable i will denote the bit // that we are currently at int i = 0; unsigned long long l = 1; while (i <= 63) { l <<= i; // If only set bit in n // is at position i if (l == n) return true ; // Get to next valid bit position i += 3; l = 1; } return false ; } // Driver code int main() { int n = 65; if (checkPowerof8(n)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function to check if n is power of 8 static boolean checkPowerof8( int n) { // Variable i will denote the bit // that we are currently at int i = 0 ; long l = 1 ; while (i <= 63 ) { l <<= i; // If only set bit in n // is at position i if (l == n) return true ; // Get to next valid bit position i += 3 ; l = 1 ; } return false ; } // Driver code public static void main (String[] args) { int n = 65 ; if (checkPowerof8(n)) System.out.println ( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by Tushil. |
Python3
# Python3 implementation of the approach # Function to check if n is power of 8 def checkPowerof8(n): # Variable i will denote the bit # that we are currently at i = 0 l = 1 while (i < = 63 ): l << = i # If only set bit in n # is at position i if (l = = n): return True # Get to next valid bit position i + = 3 l = 1 return False # Driver code if __name__ = = '__main__' : n = 65 if (checkPowerof8(n)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by math_lover |
C#
// C# implementation of the approach using System; class GFG { // Function to check if n is power of 8 static bool checkPowerof8( int n) { // Variable i will denote the bit // that we are currently at int i = 0; long l = 1; while (i <= 63) { l <<= i; // If only set bit in n // is at position i if (l == n) return true ; // Get to next valid bit position i += 3; l = 1; } return false ; } // Driver code static public void Main () { int n = 65; if (checkPowerof8(n)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by ajit. |
Javascript
<script> // Function to check if n is power of 8 function checkPowerof8( n) { // Variable i will denote the bit // that we are currently at var i = 0; var l= 1; while (i <= 63) { l<<=i; // If only set bit in n // is at position i if (l == n) return 1; // Get to next valid bit position i += 3; l = 1; } return 0; } var n = 65; if (checkPowerof8(n)) document.write( "Yes" ); else document.write( "No" ); </script> |
Output :
No
Time Complexity: O(1)
Auxiliary Space: O(1)
This article is contributed by Pranav. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!