# 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

The idea is to start from the rightmost bit and one by one check value of every bit. Following is a 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 the given number is a power of 2 or not. If given number is not a 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++

br>
 `// C++ program to find position of only set bit in a given number  ` `#include ` `using` `namespace` `std; ` ` `  `// A utility function to check whether n is a 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)? cout<<``"n = "``<

## C

 `// C program to find position of only set bit in a given number ` `#include ` ` `  `// A utility function to check whether n is a 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 number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``n = 12; ` `    ``pos = findPosition(n); ` `    ``(pos == -1)? ``printf``(````"n = %d, Invalid number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``n = 128; ` `    ``pos = findPosition(n); ` `    ``(pos == -1)? ``printf``(````"n = %d, Invalid number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find position of only set bit in a given number  ` `class` `GFG ` `{ ` `     `  `// A utility function to check whether n is a power of 2 or not.  ` `// See http://goo.gl/17Arj  ` `static` `boolean` `isPowerOfTwo(``int` `n)  ` `{ ``return` `(n > ``0` `&& ((n & (n - ``1``))==``0` `)) ? ``true``:``false``; }  ` ` `  `// Returns position of the only set bit in 'n'  ` `static` `int` `findPosition(``int` `n)  ` `{  ` `    ``if` `(!isPowerOfTwo(n))  ` `        ``return` `-``1``;  ` ` `  `    ``int` `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;  ` `    ``}  ` ` `  `    ``return` `pos;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `n = ``16``;  ` `    ``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);  ` `}  ` `} ` ` `  `// This code is contributed by mits `

## 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 `

## C#

 `// C# program to find position of only set bit in a given number  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// A utility function to check whether n is a power of 2 or not.  ` `// See http://goo.gl/17Arj  ` `static` `bool` `isPowerOfTwo(``int` `n)  ` `{ ``return` `(n > 0 && ((n & (n - 1))==0 )) ? ``true``:``false``; }  ` ` `  `// Returns position of the only set bit in 'n'  ` `static` `int` `findPosition(``int` `n)  ` `{  ` `    ``if` `(!isPowerOfTwo(n))  ` `        ``return` `-1;  ` ` `  `    ``int` `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;  ` `    ``}  ` ` `  `    ``return` `pos;  ` `}  ` ` `  `// Driver code  ` `static` `void` `Main() ` `{ ` `    ``int` `n = 16;  ` `    ``int` `pos = findPosition(n);  ` `    ``if``(pos == -1) ` `        ``Console.WriteLine(``"n = "` `+ n + ``", Invalid number"``); ` `    ``else` `        ``Console.WriteLine(``"n = "` `+ n + ``", Position "` `+ pos);  ` ` `  `    ``n = 12;  ` `    ``pos = findPosition(n);  ` `    ``if``(pos == -1) ` `        ``Console.WriteLine(``"n = "` `+ n + ``", Invalid number"``); ` `    ``else` `        ``Console.WriteLine(``"n = "` `+ n + ``", Position "` `+ pos);  ` ` `  `    ``n = 128;  ` `    ``pos = findPosition(n);  ` `    ``if``(pos == -1) ` `        ``Console.WriteLine(``"n = "` `+ n + ``", Invalid number"``); ` `    ``else` `        ``Console.WriteLine(``"n = "` `+ n + ``", Position "` `+ pos);  ` `}  ` `} ` ` `  `// This code is contributed by mits `

## PHP

 ` `

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 the position of the set bit.

## C

 `// C program to find position of only set bit in a given number ` `#include ` ` `  `// 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 number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``n = 12; ` `    ``pos = findPosition(n); ` `    ``(pos == -1)? ``printf``(````"n = %d, Invalid number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``n = 128; ` `    ``pos = findPosition(n); ` `    ``(pos == -1)? ``printf``(````"n = %d, Invalid number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, 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 `

## C#

 `// C# program to find position of only ` `// set bit in a given number ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// A utility function to check whether ` `// n is power of 2 or not ` `static` `bool` `isPowerOfTwo(``int` `n) ` `{ ` `    ``return` `n > 0 && ((n & (n-1))==0 );  ` `     `  `} ` ` `  `// Returns position of the only set bit in 'n' ` `static` `int` `findPosition(``int` `n) ` `{ ` `    ``if` `(!isPowerOfTwo(n)) ` `        ``return` `-1; ` ` `  `    ``int` `count = 0; ` ` `  `    ``// One by one move the only set bit ` `    ``// to right till it reaches end ` `    ``while` `(n > 0) ` `    ``{ ` `        ``n = n >> 1; ` ` `  `        ``// increment count of shifts ` `        ``++count; ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `n = 0; ` `    ``int` `pos = findPosition(n); ` `    ``if``(pos == -1) ` `        ``Console.WriteLine(``"n = "` `+ n + ``", Invalid number"``); ` `    ``else` `        ``Console.WriteLine(``"n = "` `+ n + ``", Position "` `+ pos); ` ` `  `    ``n = 12; ` `    ``pos = findPosition(n); ` `    ``if``(pos == -1) ` `        ``Console.WriteLine(``"n = "` `+ n + ``", Invalid number"``); ` `    ``else` `        ``Console.WriteLine(``"n = "` `+ n + ``", Position "` `+ pos); ` ` `  `    ``n = 128; ` `    ``pos = findPosition(n); ` `    ``if``(pos == -1) ` `        ``Console.WriteLine(``"n = "` `+ n + ``", Invalid number"``); ` `    ``else` `        ``Console.WriteLine(``"n = "` `+ n + ``", Position "` `+ pos); ` `} ` `} ` ` `  `// This code is cotributed by mits `

## PHP

 `> 1; ` ` `  `        ``// increment count of shifts ` `        ``++``\$count``; ` `    ``} ` ` `  `    ``return` `\$count``; ` `} ` ` `  `// Driver Code ` `\$n` `= 0; ` `\$pos` `= findPosition(``\$n``); ` `if``((``\$pos` `== -1) == true) ` `    ``echo` `"n = "``, ``\$n``, ``","``,  ` `         ``" Invalid number"``, ````" "````; ` `else` `    ``echo` `"n = "``, ``\$n``, ``","``, ` `         ``" Position "``, ``\$pos``, ````" "````; ` ` `  `\$n` `= 12; ` `\$pos` `= findPosition(``\$n``); ` `if` `((``\$pos` `== -1) == true) ` `        ``echo` `"n = "``, ``\$n``, ``","``, ` `             ``" Invalid number"``, ````" "````; ` `else` `        ``echo` `"n = "``, ``\$n``,  ` `             ``" Position "``, ``\$pos``, ````" "````; ` ` `  `\$n` `= 128; ` `\$pos` `= findPosition(``\$n``); ` `    ``if``((``\$pos` `== -1) == true) ` `        ``echo` `"n = "``, ``\$n``, ``","``,  ` `             ``" Invalid number"``,````" "````; ` `else` `        ``echo` `"n = "``, ``\$n``, ``","``,  ` `             ``" Position "``, ``\$pos``, ````" "````; ` `         `  `// 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 ` ` `  `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 number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``n = 12; ` `    ``pos = findPosition(n); ` `    ``(pos == -1)? ``printf``(````"n = %d, Invalid number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, n, pos); ` ` `  `    ``n = 128; ` `    ``pos = findPosition(n); ` `    ``(pos == -1)? ``printf``(````"n = %d, Invalid number "````, n): ` `                 ``printf``(````"n = %d, Position %d "````, 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

 ` 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"``, ````" "````; ` `else` `        ``echo` `"n = "``,``\$n``, ``","``,  ` `             ``" Position n"``, ``\$pos``,````" "````; ` ` `  `\$n` `= 12; ` `\$pos` `= findPosition(``\$n``); ` `if``((``\$pos` `== -1) == true) ` `            ``echo` `"n = "``, ``\$n``,``","` `,  ` `                 ``" Invalid number"``,````" "````; ` `        ``else` `            ``echo` `"n ="``, ``\$n``,``","``,  ` `                 ``" Position"``, ``\$pos``, ````" "````; ` ` `  `// Driver Code ` `\$n` `= 128; ` `\$pos` `= findPosition(``\$n``); ` `if``((``\$pos` `== -1) == true) ` `        ``echo` `"n = "``, ``\$n``, ``","``,  ` `             ``" Invalid number"``,````" "````; ` `else` `        ``echo` `"n = "``, ``\$n``, ``","``,  ` `             ``" Position "``,``\$pos``, ````" "````; ` `         `  `// 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.

This article is attributed to GeeksforGeeks.org

## tags:

Bit Magic Microsoft Microsoft Bit Magic

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