Find position of the only set bit | Java program to find position of only set bit in a given number ; A utility function to check whether n is power of 2 or not ; Returns position of the only set bit in ' n ' ; One by one move the only set bit to right till it reaches end ; increment count of shifts ; Driver code
class GFG { static boolean isPowerOfTwo ( int n ) { return n > 0 && ( ( n & ( n - 1 ) ) == 0 ) ; } static int findPosition ( int n ) { if ( ! isPowerOfTwo ( n ) ) return - 1 ; int count = 0 ; while ( n > 0 ) { n = n >> 1 ; ++ count ; } return count ; } public static void main ( String [ ] args ) { int n = 0 ; int pos = findPosition ( n ) ; if ( pos == - 1 ) System . out . println ( " n ▁ = ▁ " + n + " , ▁ Invalid ▁ number " ) ; else System . out . println ( " n ▁ = ▁ " + n + " , ▁ Position ▁ " + pos ) ; n = 12 ; pos = findPosition ( n ) ; if ( pos == - 1 ) System . out . println ( " n ▁ = ▁ " + n + " , ▁ Invalid ▁ number " ) ; else System . out . println ( " n ▁ = ▁ " + n + " , ▁ Position ▁ " + pos ) ; n = 128 ; pos = findPosition ( n ) ; if ( pos == - 1 ) System . out . println ( " n ▁ = ▁ " + n + " , ▁ Invalid ▁ number " ) ; else System . out . println ( " n ▁ = ▁ " + n + " , ▁ Position ▁ " + pos ) ; } }