# XOR of two numbers after making length of their binary representations equal

Given two numbers say a and b. Print their XOR after making the lengths of their binary representation equal by adding trailing zeros to the binary representation of smaller one.

Examples :

```Input : a = 13, b = 5
Output : 7
Explanation : Binary representation of 13 is 1101 and
of 5 is 101. As the length of "101" is smaller,
so add a '0' to it making it "1010', to make
the length of binary representations equal.
XOR of 1010 and 1101 gives 0111 which is 7.

Input : a = 7, b = 5
Output : 2
Explanation : Since the length of binary representations
of 7 i.e, 111 and 5 i.e, 101 are same, hence simply
print XOR of a and b.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : Count the number of bits in binary representation of smaller number out of a and b. If the number of bits in smaller number(say a) exceeds to that of larger number(say b), then apply left shift to the smaller number by the number of exceeding bits, i.e, a = a<<(exceeding bits). After applying left shift, trailing zeroes will be added at the end of binary representation of smaller number to make the number of bits in binary representation of both the numbers equal. XOR both the binary representations to get the final result.

Below is the implementation of above method :

## C++

 `// C++ implementation to return  ` `// XOR of two numbers after making ` `// length of their binary representation same ` `#include ` `using` `namespace` `std; ` ` `  `// function to count the number  ` `// of bits in binary representation  ` `// of an integer ` `int` `count(``int` `n) ` `{ ` `    ``// initialize count ` `    ``int` `c = 0; ` `     `  `    ``// count till n is non zero ` `    ``while` `(n) ` `    ``{ ` `        ``c++; ` `         `  `        ``// right shift by 1  ` `        ``// i.e, divide by 2 ` `        ``n = n>>1; ` `    ``} ` `    ``return` `c; ` `} ` ` `  `// function to calculate the xor of ` `// two numbers by adding trailing  ` `// zeros to the number having less number  ` `// of bits in its binary representation. ` `int` `XOR(``int` `a, ``int` `b) ` `{ ` `    ``// stores the minimum and maximum  ` `    ``int` `c = min(a,b); ` `    ``int` `d = max(a,b); ` `     `  `    ``// left shift if the number of bits ` `    ``// are less in binary representation ` `    ``if` `(count(c) < count(d)) ` `       ``c = c << ( count(d) - count(c) );  ` `     `  `    ``return` `(c^d);  ` `} ` ` `  `// driver code to check the above function  ` `int` `main() ` `{    ` `    ``int` `a = 13, b = 5; ` `    ``cout << XOR(a,b);     ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to return  ` `// XOR of two numbers after making ` `// length of their binary representation same ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `    ``// function to count the number  ` `    ``// of bits in binary representation  ` `    ``// of an integer ` `    ``static` `int` `count(``int` `n) ` `    ``{ ` `        ``// initialize count ` `        ``int` `c = ``0``; ` `         `  `        ``// count till n is non zero ` `        ``while` `(n != ``0``) ` `        ``{ ` `            ``c++; ` `             `  `            ``// right shift by 1  ` `            ``// i.e, divide by 2 ` `            ``n = n >> ``1``; ` `        ``} ` `        ``return` `c; ` `    ``} ` `     `  `    ``// function to calculate the xor of ` `    ``// two numbers by adding trailing  ` `    ``// zeros to the number having less number  ` `    ``// of bits in its binary representation. ` `    ``static` `int` `XOR(``int` `a, ``int` `b) ` `    ``{ ` `        ``// stores the minimum and maximum  ` `        ``int` `c = Math.min(a, b); ` `        ``int` `d = Math.max(a, b); ` `         `  `        ``// left shift if the number of bits ` `        ``// are less in binary representation ` `        ``if` `(count(c) < count(d)) ` `        ``c = c << ( count(d) - count(c) );  ` `         `  `        ``return` `(c ^ d);  ` `    ``} ` `     `  `    ``// driver code to check the above function  ` `    ``public` `static` `void` `main(String args[]) ` `    ``{  ` `        ``int` `a = ``13``, b = ``5``; ` `        ``System.out.println(XOR(a, b)); ` `    ``} ` `} ` ` `  `// This code is contributed by Nikita Tiwari. `

## Python3

 `# Python 3 implementation to return XOR ` `# of two numbers after making length ` `# of their binary representation same ` ` `  `# Function to count the number of bits ` `# in binary representation of an integer ` `def` `count(n) : ` `     `  `    ``# initialize count ` `    ``c ``=` `0` `     `  `    ``# count till n is non zero ` `    ``while` `(n !``=` `0``) : ` `        ``c ``+``=` `1` `         `  `        ``# right shift by 1  ` `        ``# i.e, divide by 2 ` `        ``n ``=` `n >> ``1` `         `  `    ``return` `c ` `     `  `# Function to calculate the xor of ` `# two numbers by adding trailing  ` `# zeros to the number having less number  ` `# of bits in its binary representation. ` `def` `XOR(a, b) : ` `     `  `    ``# stores the minimum and maximum  ` `    ``c ``=` `min``(a, b) ` `    ``d ``=` `max``(a, b) ` `     `  `    ``# left shift if the number of bits ` `    ``# are less in binary representation ` `    ``if` `(count(c) < count(d)) : ` `        ``c ``=` `c << ( count(d) ``-` `count(c) )  ` `     `  `    ``return` `(c^d) ` ` `  `# Driver Code ` `a ``=` `13``; b ``=` `5` `print``(XOR(a, b)) ` ` `  ` `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# implementation to return XOR of two  ` `// numbers after making length of their  ` `// binary representation same ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// function to count the number  ` `    ``// of bits in binary representation  ` `    ``// of an integer ` `    ``static` `int` `count(``int` `n) ` `    ``{ ` `         `  `        ``// initialize count ` `        ``int` `c = 0; ` `         `  `        ``// count till n is non zero ` `        ``while` `(n != 0) ` `        ``{ ` `             `  `            ``c++; ` `             `  `            ``// right shift by 1  ` `            ``// i.e, divide by 2 ` `            ``n = n >> 1; ` `        ``} ` `         `  `        ``return` `c; ` `    ``} ` `     `  `    ``// function to calculate the xor of ` `    ``// two numbers by adding trailing  ` `    ``// zeros to the number having less number  ` `    ``// of bits in its binary representation. ` `    ``static` `int` `XOR(``int` `a, ``int` `b) ` `    ``{ ` `         `  `        ``// stores the minimum and maximum  ` `        ``int` `c = Math.Min(a, b); ` `        ``int` `d = Math.Max(a, b); ` `         `  `        ``// left shift if the number of bits ` `        ``// are less in binary representation ` `        ``if` `(count(c) < count(d)) ` `        ``c = c << ( count(d) - count(c) );  ` `         `  `        ``return` `(c ^ d);  ` `    ``} ` `     `  `    ``// driver code to check the above function  ` `    ``public` `static` `void` `Main() ` `    ``{  ` `        ``int` `a = 13, b = 5; ` `         `  `        ``Console.WriteLine(XOR(a, b)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `>1; ` `    ``} ` `    ``return` `\$c``; ` `} ` ` `  `// function to calculate the xor of ` `// two numbers by adding trailing  ` `// zeros to the number having less number  ` `// of bits in its binary representation. ` `function` `XOR1(``\$a``, ``\$b``) ` `{ ` `     `  `    ``// stores the minimum  ` `    ``// and maximum  ` `    ``\$c` `= min(``\$a``,``\$b``); ` `    ``\$d` `= max(``\$a``,``\$b``); ` `     `  `    ``// left shift if the number of bits ` `    ``// are less in binary representation ` `    ``if` `(count1(``\$c``) < count1(``\$d``)) ` `    ``\$c` `= ``\$c` `<< ( count1(``\$d``) - count1(``\$c``) );  ` `     `  `    ``return` `(``\$c``^``\$d``);  ` `} ` ` `  `    ``// Driver Code  ` `    ``\$a` `= 13; ` `    ``\$b` `= 5; ` `    ``echo` `XOR1(``\$a``, ``\$b``);  ` ` `  `// This code is contributed by mits  ` `?> `

Output :

```7
```

My Personal Notes arrow_drop_up Striver(underscore)79 at Codechef and codeforces D

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