Skip to content
Related Articles
Find position of left most dis-similar bit for two numbers
• Last Updated : 07 Apr, 2021

Given two numbers n1 and n2. The task is to find the position of first mismatching bit in the binary representation of the two numbers from left. We need to find this bit after making lengths of binary representations of both numbers same. We make lengths same by appending 0s in the smaller number.

Note

• For Example: n1 = 1, n2 = 7. Bitwise representation of n1 and n4 will be “1” and “111” respectively. Append two zeros to n1 to make it 100.
• Print zero if n1 is equal to n2.

Examples:

```Input: n1 = 12, n2 = 34
Output: 2
Binary representation of 12 is 1100 and of 34 is 100010.
First make both representations of the
same length by appending 0s.
So the first representation now becomes 11000.
The second bit is the different bit.

Input: n1 = 1, n2 = 2
Output: 2```

For finding the position of leftmost dis-similar bit among bit representaion of two numbers, either bit by bit comparison can be done or a derived formula can be used. Although the time complexity for both is same.
For finding the left most dis-similar bit first of all equalize the bit length of both numbers by multiply the smaller one with pow(2, bit-length Difference). After making bit-length equal take XOR of both number. Now, leftmost dis-similar bit is clearly reflected in XOR value. Subtracting the bit-length of XOR value from bit-length of given number plus 1 is the position of leftmost dis-similar bit can be concluded.

Algorithm:

1. Find bit length of n1 & n2.
2. Equalize bit-length of both number by placing zero to right of smaller number (same as multiplying smaller one by pow(2, bit-length Difference) )
3. Take XOR of both number
4. Difference of bit-length of any number and bit-length of XOR value is required answer plus 1

Below is the implementation of the above approach:

## C++

 `// C++ program to find the leftmost``// position of first dis-similar bit``#include ``using` `namespace` `std;` `// Function to find first dis-similar bit``int` `bitPos(``int` `n1, ``int` `n2)``{``    ``// return zero for equal number``    ``if` `(n1 == n2)``        ``return` `0;` `    ``/** find the 1st dis-similar bit **/``    ``// count bit length of n1 and n2``    ``int` `bitCount1 = ``floor``(log2(n1)) + 1;``    ``int` `bitCount2 = ``floor``(log2(n2)) + 1;` `    ``// find bit difference and maxBit``    ``int` `bitDiff = ``abs``(bitCount1 - bitCount2);``    ``int` `maxBitCount = max(bitCount1, bitCount2);` `    ``if` `(bitCount1 > bitCount2) {``        ``n2 = n2 * ``pow``(2, bitDiff);``    ``}``    ``else` `{``        ``n1 = n1 * ``pow``(2, bitDiff);``    ``}` `    ``int` `xorValue = n1 ^ n2;``    ``int` `bitCountXorValue = ``floor``(log2(xorValue)) + 1;``    ``int` `disSimilarBitPosition = maxBitCount -``                                  ``bitCountXorValue + 1;` `    ``return` `disSimilarBitPosition;``}` `// Driver program``int` `main()``{``    ``int` `n1 = 53, n2 = 55;``    ``cout << bitPos(n1, n2);``    ``return` `0;``}`

## Java

 `// Java program to find the leftmost position of``// first dis-similar bit` `import` `java.io.*;` `class` `GFG {``        ` `// Function to find first dis-similar bit``static` `int` `bitPos(``int` `n1, ``int` `n2)``{``    ``// return zero for equal number``    ``if` `(n1 == n2)``        ``return` `0``;` `    ``/** find the 1st dis-similar bit **/``    ``// count bit length of n1 and n2``    ``int` `bitCount1 = (``int``)Math.floor(Math.log(n1) /``                                    ``Math.log(``2``)) + ``1``;``    ``int` `bitCount2 = (``int``)Math.floor(Math.log(n2) /``                                    ``Math.log(``2``)) + ``1``;` `    ``// find bit difference and maxBit``    ``int` `bitDiff = Math.abs(bitCount1 - bitCount2);``    ``int` `maxBitCount = Math.max(bitCount1,``                            ``bitCount2);` `    ``if` `(bitCount1 > bitCount2)``    ``{``        ``n2 = n2 * (``int``)Math.pow(``2``, bitDiff);``    ``}``    ``else``    ``{``        ``n1 = n1 * (``int``)Math.pow(``2``, bitDiff);``    ``}` `    ``int` `xorValue = n1 ^ n2;``    ``int` `bitCountXorValue = (``int``)Math.floor(Math.log(xorValue) /``                                        ``Math.log(``2``)) + ``1``;``    ``int` `disSimilarBitPosition = maxBitCount -``                                ``bitCountXorValue + ``1``;` `    ``return` `disSimilarBitPosition;``}` `// Driver Code``    ``public` `static` `void` `main (String[] args) {` `        ``int` `n1 = ``53``, n2 = ``55``;``        ``System.out.println(bitPos(n1, n2));``}``}``// This code is contributed by ajit`

## Python3

 `# Python3 program to Find the leftmost``# position of first dis-similar bit` `# from math lib import floor()``# and log2()``from` `math ``import` `floor, log2` `# Function to find first``# dis-similar bit``def` `bitPos(n1, n2) :``    ` `    ``# return zero for equal number``    ``if` `n1 ``=``=` `n2 :``        ``return` `0``    ` `    ``# find the 1st dis-similar bit``    ``# count bit length of n1 and n``    ``bitCount1 ``=` `floor(log2(n1)) ``+` `1``    ``bitCount2 ``=` `floor(log2(n2)) ``+` `1``    ` `    ``# find bit difference and maxBit``    ``bitDiff ``=` `abs``(bitCount1 ``-` `bitCount2)``    ``maxBitCount ``=` `max``(bitCount1, bitCount2)``    ` `    ``if` `(bitCount1 > bitCount2) :``        ` `        ``n2 ``*``=` `pow``(``2``, bitDiff)``    ` `    ``else` `:``        ` `        ``n1 ``*``=` `pow``(``2``, bitDiff)``        ` `    ``xorValue ``=` `n1 ^ n2``    ``bitCountXorValue ``=` `floor(log2(xorValue)) ``+` `1``    ``disSimilarBitPosition ``=` `(maxBitCount ``-``                             ``bitCountXorValue ``+` `1``)``    ` `    ``return` `disSimilarBitPosition``    ` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``n1, n2 ``=` `53``, ``55``    ``print``(bitPos(n1, n2))``    ` `# This code is contributed by Ryuga`

## C#

 `// C# to find the leftmost position of``// first dis-similar bit``using` `System;` `class` `GFG``{``    ` `// Function to find first dis-similar bit``static` `int` `bitPos(``int` `n1, ``int` `n2)``{``    ``// return zero for equal number``    ``if` `(n1 == n2)``        ``return` `0;` `    ``/** find the 1st dis-similar bit **/``    ``// count bit length of n1 and n2``    ``int` `bitCount1 = (``int``)Math.Floor(Math.Log(n1) /``                                    ``Math.Log(2)) + 1;``    ``int` `bitCount2 = (``int``)Math.Floor(Math.Log(n2) /``                                    ``Math.Log(2)) + 1;` `    ``// find bit difference and maxBit``    ``int` `bitDiff = Math.Abs(bitCount1 - bitCount2);``    ``int` `maxBitCount = Math.Max(bitCount1,``                               ``bitCount2);` `    ``if` `(bitCount1 > bitCount2)``    ``{``        ``n2 = n2 * (``int``)Math.Pow(2, bitDiff);``    ``}``    ``else``    ``{``        ``n1 = n1 * (``int``)Math.Pow(2, bitDiff);``    ``}` `    ``int` `xorValue = n1 ^ n2;``    ``int` `bitCountXorValue = (``int``)Math.Floor(Math.Log(xorValue) /``                                           ``Math.Log(2)) + 1;``    ``int` `disSimilarBitPosition = maxBitCount -``                                ``bitCountXorValue + 1;` `    ``return` `disSimilarBitPosition;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n1 = 53, n2 = 55;``    ``Console.Write(bitPos(n1, n2));``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

 ` ``\$bitCount2``)``    ``{``        ``\$n2` `= ``\$n2` `* pow(2, ``\$bitDiff``);``    ``}``    ``else` `{``        ``\$n1` `= ``\$n1` `* pow(2, ``\$bitDiff``);``    ``}` `    ``\$xorValue` `= ``\$n1` `^ ``\$n2``;``    ``\$bitCountXorValue` `= ``floor``(log(``\$xorValue``, 2)) + 1;``    ``\$disSimilarBitPosition` `= ``\$maxBitCount` `-``                             ``\$bitCountXorValue` `+ 1;` `    ``return` `\$disSimilarBitPosition``;``}` `// Driver Code``\$n1` `= 53;``\$n2` `= 55;``echo` `bitPos(``\$n1``, ``\$n2``);``    ` `// This code is contributed by ajit``?>`

## Javascript

 ``
Output:
`5`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up