Find position of the only set bit

Given a number having only one ‘1’ and all other ’0’s in its binary representation, find position of the only set bit. Source: Microsoft Interview | 18

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to start from rightmost bit and one by one check value of every bit. Following is detailed algorithm.

1) If number is power of two then and then only its binary representation contains only one ‘1’. That’s why check whether given number is power of 2 or not. If given number is not power of 2, then print error message and exit.

2) Initialize two variables; i = 1 (for looping) and pos = 1 (to find position of set bit)

3) Inside loop, do bitwise AND of i and number ‘N’. If value of this operation is true, then “pos” bit is set, so break the loop and return position. Otherwise, increment “pos” by 1 and left shift i by 1 and repeat the procedure.

C

// C program to find position of only set bit in a given number 
#include <stdio.h> 
  
// A utility function to check whether n is power of 2 or not. See http://goo.gl/17Arj 
int isPowerOfTwo(unsigned n) 
{  return n && (! (n & (n-1)) ); } 
  
// Returns position of the only set bit in 'n' 
int findPosition(unsigned n) 
{ 
    if (!isPowerOfTwo(n)) 
        return -1; 
  
    unsigned i = 1, pos = 1; 
  
    // Iterate through bits of n till we find a set bit 
    // i&n will be non-zero only when 'i' and 'n' have a set bit 
    // at same position 
    while (!(i & n)) 
    { 
        // Unset current bit and set the next bit in 'i' 
        i = i << 1; 
  
        // increment position 
        ++pos; 
    } 
  
    return pos; 
} 
  
// Driver program to test above function 
int main(void) 
{ 
    int n = 16; 
    int pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    n = 12; 
    pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    n = 128; 
    pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    return 0; 
} 

Python3

# Python3 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. 
def isPowerOfTwo(n): 
    return (True if(n > 0 and 
                   ((n & (n - 1)) > 0))  
                 else False);  
      
# Returns position of the 
# only set bit in 'n' 
def findPosition(n): 
    if (isPowerOfTwo(n) == True): 
        return -1; 
  
    i = 1; 
    pos = 1; 
  
    # Iterate through bits of n  
    # till we find a set bit i&n 
    # will be non-zero only when 
    # 'i' and 'n' have a set bit 
    # at same position 
    while ((i & n) == 0): 
          
        # Unset current bit and  
        # set the next bit in 'i' 
        i = i << 1; 
  
        # increment position 
        pos += 1; 
  
    return pos; 
  
# Driver Code 
n = 16; 
pos = findPosition(n); 
if (pos == -1): 
    print("n =", n, ", Invalid number"); 
else: 
    print("n =", n, ", Position ", pos); 
  
n = 12; 
pos = findPosition(n); 
if (pos == -1): 
    print("n =", n, ", Invalid number"); 
else: 
    print("n =", n, ", Position ", pos); 
  
n = 128; 
pos = findPosition(n); 
if (pos == -1): 
    print("n =", n, ", Invalid number"); 
else: 
    print("n =", n," , Position ", pos); 
  
# This code is contributed by mits 

PHP

<?php 
// PHP 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. See http://goo.gl/17Arj 
function isPowerOfTwo($n) 
{ 
    return $n && (!($n & ($n - 1)));  
      
} 
  
// Returns position of the 
// only set bit in 'n' 
function findPosition($n) 
{ 
    if (!isPowerOfTwo($n)) 
        return -1; 
  
    $i = 1; 
    $pos = 1; 
  
    // Iterate through bits of n  
    // till we find a set bit i&n 
    // will be non-zero only when 
    // 'i' and 'n' have a set bit 
    // at same position 
    while (!($i & $n)) 
    { 
        // Unset current bit and  
        // set the next bit in 'i' 
        $i = $i << 1; 
  
        // increment position 
        ++$pos; 
    } 
  
    return $pos; 
} 
  
// Driver Code 
$n = 16; 
$pos = findPosition($n); 
if (($pos == -1) == true) 
        echo "n =", $n, "," , 
             " Invalid number" , "\n"; 
else
        echo "n = ", $n, ",", 
             " Position ", $pos, "\n"; 
  
$n = 12; 
$pos = findPosition($n); 
if(($pos == -1) == true)      
        echo "n = ", $n, ",",  
             " Invalid number", "\n"; 
else
        echo "n =", $n, ",",  
             " Position ", $pos, "\n"; 
$n = 128; 
$pos = findPosition($n); 
if(($pos == -1) == true)      
        echo "n =", $n, ",", 
             " Invalid number", "\n"; 
else
        echo "n = ", $n, ",", 
             " Position ", $pos, "\n"; 
  
// This code is contributed by ajit 
?> 

Output :

n = 16, Position 5
n = 12, Invalid number
n = 128, Position 8

Following is another method for this problem. The idea is to one by one right shift the set bit of given number ‘n’ until ‘n’ becomes 0. Count how many times we shifted to make ‘n’ zero. The final count is position of the set bit.

C

// C program to find position of only set bit in a given number 
#include <stdio.h> 
  
// A utility function to check whether n is power of 2 or not 
int isPowerOfTwo(unsigned n) 
{  return n && (! (n & (n-1)) ); } 
  
// Returns position of the only set bit in 'n' 
int findPosition(unsigned n) 
{ 
    if (!isPowerOfTwo(n)) 
        return -1; 
  
    unsigned count = 0; 
  
    // One by one move the only set bit to right till it reaches end 
    while (n) 
    { 
        n = n >> 1; 
  
        // increment count of shifts 
        ++count; 
    } 
  
    return count; 
} 
  
// Driver program to test above function 
int main(void) 
{ 
    int n = 0; 
    int pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    n = 12; 
    pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    n = 128; 
    pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    return 0; 
} 

Python3

# Python 3 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  
def isPowerOfTwo(n) : 
  
    return (n and ( not (n & (n-1)))) 
  
# Returns position of the only set bit in 'n' 
def findPosition(n) : 
  
    if not isPowerOfTwo(n) : 
        return -1
  
    count = 0
  
    # One by one move the only set bit to  
    # right till it reaches end 
    while (n) : 
          
        n = n >> 1
  
        # increment count of shifts  
        count += 1
  
    return count 
  
  
# Driver program to test above function  
if __name__ == "__main__" : 
    n = 0
    pos = findPosition(n) 
  
    if pos == -1 : 
        print("n =",n,"Invalid number") 
    else : 
        print("n =",n,"Position",pos) 
  
    n = 12
    pos = findPosition(n) 
  
    if pos == -1 : 
        print("n =",n,"Invalid number") 
    else : 
        print("n =",n,"Position",pos) 
  
    n = 128
    pos = findPosition(n) 
  
    if pos == -1 : 
        print("n =",n,"Invalid number") 
    else : 
        print("n =",n,"Position",pos) 
     
# This code is contributed by ANKITRAI1 

PHP

<?php 
// PHP 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 
function isPowerOfTwo($n) 
{  
    return $n && (! ($n & ($n - 1)));  
} 
  
// Returns position of the 
// only set bit in 'n' 
function findPosition($n) 
{ 
    if (!isPowerOfTwo($n)) 
        return -1; 
  
    $count = 0; 
  
    // One by one move the only set 
    // bit to right till it reaches end 
    while ($n) 
    { 
        $n = $n >> 1; 
  
        // increment count of shifts 
        ++$count; 
    } 
  
    return $count; 
} 
  
// Driver Code 
$n = 0; 
$pos = findPosition($n); 
if(($pos == -1) == true) 
    echo "n = ", $n, ",",  
         " Invalid number", "\n"; 
else
    echo "n = ", $n, ",", 
         " Position ", $pos, "\n"; 
  
$n = 12; 
$pos = findPosition($n); 
if (($pos == -1) == true) 
        echo "n = ", $n, ",", 
             " Invalid number", "\n"; 
else
        echo "n = ", $n,  
             " Position ", $pos, "\n"; 
  
$n = 128; 
$pos = findPosition($n); 
    if(($pos == -1) == true) 
        echo "n = ", $n, ",",  
             " Invalid number","\n"; 
else
        echo "n = ", $n, ",",  
             " Position ", $pos, "\n"; 
          
// This code is contributed by ajit 
?> 

Output :

n = 0, Invalid number
n = 12, Invalid number
n = 128, Position 8

We can also use log base 2 to find the position. Thanks to Arunkumar for suggesting this solution.

C

#include <stdio.h> 
  
unsigned int Log2n(unsigned int n) 
{ 
   return (n > 1)? 1 + Log2n(n/2): 0; 
} 
  
int isPowerOfTwo(unsigned n) 
{ 
    return n && (! (n & (n-1)) ); 
} 
  
int findPosition(unsigned n) 
{ 
    if (!isPowerOfTwo(n)) 
        return -1; 
    return Log2n(n) + 1; 
} 
  
// Driver program to test above function 
int main(void) 
{ 
    int n = 0; 
    int pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    n = 12; 
    pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    n = 128; 
    pos = findPosition(n); 
    (pos == -1)? printf("n = %d, Invalid numbern", n): 
                 printf("n = %d, Position %d n", n, pos); 
  
    return 0; 
} 

Python3

# Python program to find position 
# of only set bit in a given number 
  
def Log2n(n): 
    if (n > 1): 
        return (1 + Log2n(n/2)) 
    else: 
        return 0
      
# A utility function to check 
# whether n is power of 2 or not     
def isPowerOfTwo(n): 
    return n and (not (n & (n-1)) ) 
      
def findPosition(n): 
    if (not isPowerOfTwo(n)): 
        return -1
    return Log2n(n) + 1
      
# Driver program to test above function 
  
n = 0
pos = findPosition(n) 
if(pos == -1): 
    print("n =", n, ", Invalid number") 
else: 
    print("n = ", n, ", Position ", pos) 
   
n = 12
pos = findPosition(n) 
if(pos == -1): 
    print("n =", n, ", Invalid number") 
else: 
    print("n = ", n, ", Position ", pos) 
n = 128
pos = findPosition(n) 
if(pos == -1): 
    print("n = ", n, ", Invalid number") 
else: 
    print("n = ", n, ", Position ", pos) 
   
# This code is contributed 
# by Sumit Sudhakar 

PHP

<?php 
// PHP program to find position 
// of only set bit in a given number 
function Log2n($n) 
{ 
return ($n > 1) ? 1 +  
  Log2n($n / 2) : 0; 
} 
  
function isPowerOfTwo($n) 
{ 
    return $n && (! ($n &  
                    ($n - 1))); 
} 
  
function findPosition($n) 
{ 
    if (!isPowerOfTwo($n)) 
        return -1; 
    return Log2n($n) + 1; 
} 
  
// Driver Code 
$n = 0; 
$pos = findPosition($n); 
if(($pos == -1) == true) 
        echo "n =", $n, ", " ,  
             " Invalid number", "\n"; 
else
        echo "n = ",$n, ",",  
             " Position n", $pos,"\n"; 
  
$n = 12; 
$pos = findPosition($n); 
if(($pos == -1) == true) 
            echo "n = ", $n,"," ,  
                 " Invalid number","\n"; 
        else
            echo "n =", $n,",",  
                 " Position", $pos, "\n"; 
  
// Driver Code 
$n = 128; 
$pos = findPosition($n); 
if(($pos == -1) == true) 
        echo "n = ", $n, ",",  
             " Invalid number","\n"; 
else
        echo "n = ", $n, ",",  
             " Position ",$pos, "\n"; 
          
// This code is contributed by aj_36 
?> 

Output :

n = 0, Invalid number
n = 12, Invalid number
n = 128, Position 8

This article is compiled by Narendra Kangralkar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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