# Check if bits in range L to R of two numbers are complement of each other or not

• Last Updated : 07 Apr, 2021

Given two non-negative numbers a and b and two values l and r. The problem is to check whether all bits at corresponding positions in the range l to r in both the given numbers are complement of each other or not.
The bits are numbered from right to left, i.e., the least significant bit is considered to be at first position.
Examples

```Input: a = 10, b = 5
l = 1, r = 3
Output: Yes
(10)10 = (1010)2
(5)10 = (101)2 = (0101)2
All the bits in the range 1 to 3 are complement of each other.

Input: a = 21, b = 13
l = 2, r = 4
Output: No
(21)10 = (10101)2
(13)10 = (1101)2 = (1101)2
All the bits in the range 2 to 4 are not complement of each other.```

Approach: Below are the steps to solve the problem

• Calculate xor_value = a ^ b.
• Check whether all the bits are set or not in the range l to r in xor_value. Refer this post.

Below is the implementation of the above approach.

## C++

 `// C++ implementation to check ` `// whether all the bits in the given range` `// of two numbers are complement of each other` `#include ` `using` `namespace` `std;`   `// function to check whether all the bits` `// are set in the given range or not` `bool` `allBitsSetInTheGivenRange(unsigned ``int` `n,` `                               ``unsigned ``int` `l, ` `                               ``unsigned ``int` `r)` `{` `    ``// calculating a number 'num' having 'r'` `    ``// number of bits and bits in the range l` `    ``// to r are the only set bits` `    ``int` `num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);`   `    ``// new number which will only have one or more` `    ``// set bits in the range l to r and nowhere else` `    ``int` `new_num = n & num;`   `    ``// if both are equal, then all bits are set` `    ``// in the given range` `    ``if` `(num == new_num)` `        ``return` `true``;`   `    ``// else all bits are not set` `    ``return` `false``;` `}`   `// function to check whether all the bits in the given range` `// of two numbers are complement of each other` `bool` `bitsAreComplement(unsigned ``int` `a, unsigned ``int` `b,` `                       ``unsigned ``int` `l, unsigned ``int` `r)` `{` `    ``unsigned ``int` `xor_value = a ^ b;` `    ``return` `allBitsSetInTheGivenRange(xor_value, l, r);` `}`   `// Driver Code` `int` `main()` `{` `    ``unsigned ``int` `a = 10, b = 5;` `    ``unsigned ``int` `l = 1, r = 3;`   `    ``if` `(bitsAreComplement(a, b, l, r))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;`   `    ``return` `0;` `}`

## Java

 `// Java implementation to check ` `// whether all the bits in the ` `// given range of two numbers ` `// are complement of each other` `class` `GFG` `{` `// function to check whether ` `// all the bits are set in ` `// the given range or not` `static` `boolean` `allBitsSetInTheGivenRange(``int` `n,` `                                         ``int` `l, ``int` `r)` `{` `    ``// calculating a number 'num'` `    ``// having 'r' number of bits ` `    ``// and bits in the range l` `    ``// to r are the only set bits` `    ``int` `num = ((``1` `<< r) - ``1``) ^ ` `              ``((``1` `<< (l - ``1``)) - ``1``);`   `    ``// new number which will only ` `    ``// have one or more set bits ` `    ``// in the range l to r and ` `    ``// nowhere else` `    ``int` `new_num = n & num;`   `    ``// if both are equal, ` `    ``// then all bits are set` `    ``// in the given range` `    ``if` `(num == new_num)` `        ``return` `true``;`   `    ``// else all bits are not set` `    ``return` `false``;` `}`   `// function to check whether all ` `// the bits in the given range` `// of two numbers are complement ` `// of each other` `static` `boolean` `bitsAreComplement(``int` `a, ``int` `b,` `                                 ``int` `l, ``int` `r)` `{` `    ``int` `xor_value = a ^ b;` `    ``return` `allBitsSetInTheGivenRange(xor_value, l, r);` `}`   `// Driver Code` `public` `static` `void` `main(String []args)` `{` `    ``int` `a = ``10``, b = ``5``;` `    ``int` `l = ``1``, r = ``3``;`   `    ``if` `(bitsAreComplement(a, b, l, r))` `        ``System.out.println(``"Yes"``);` `    ``else` `        ``System.out.println(``"No"``);` `}` `}`   `// This code is contributed by Smitha`

## Python 3

 `# Python 3 implementation to check whether ` `# all the bits in the given range of two ` `# numbers are complement of each other`   `# function to check whether all the bits` `# are set in the given range or not` `def` `allBitsSetInTheGivenRange(n, l, r):` `    `  `    ``# calculating a number 'num' having 'r'` `    ``# number of bits and bits in the range l` `    ``# to r are the only set bits` `    ``num ``=` `((``1` `<< r) ``-` `1``) ^ ((``1` `<< (l ``-` `1``)) ``-` `1``)`   `    ``# new number which will only have one ` `    ``# or more set bits in the range l to r` `    ``# and nowhere else` `    ``new_num ``=` `n & num`   `    ``# if both are equal, then all bits ` `    ``# are set in the given range` `    ``if` `(num ``=``=` `new_num):` `        ``return` `True`   `    ``# else all bits are not set` `    ``return` `False`   `# function to check whether all the bits ` `# in the given range of two numbers are ` `# complement of each other` `def` `bitsAreComplement(a, b, l, r):` `    ``xor_value ``=` `a ^ b` `    ``return` `allBitsSetInTheGivenRange(xor_value, l, r)`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``a ``=` `10` `    ``b ``=` `5` `    ``l ``=` `1` `    ``r ``=` `3`   `    ``if` `(bitsAreComplement(a, b, l, r)):` `        ``print``(``"Yes"``)` `    ``else``:` `        ``print``(``"No"``)`   `# This code is contributed by ita_c`

## C#

 `// C# implementation to check ` `// whether all the bits in the ` `// given range of two numbers ` `// are complement of each other ` `using` `System;`   `class` `GFG` `{` `// function to check whether ` `// all the bits are set in ` `// the given range or not ` `static` `bool` `allBitsSetInTheGivenRange(``int` `n, ``int` `l,` `                                      ``int` `r) ` `{ ` `    ``// calculating a number 'num' ` `    ``// having 'r' number of bits ` `    ``// and bits in the range l ` `    ``// to r are the only set bits ` `    ``int` `num = ((1 << r) - 1) ^ ` `              ``((1 << (l - 1)) - 1); `   `    ``// new number which will only ` `    ``// have one or more set bits ` `    ``// in the range l to r and ` `    ``// nowhere else ` `    ``int` `new_num = n & num; `   `    ``// if both are equal, ` `    ``// then all bits are set ` `    ``// in the given range ` `    ``if` `(num == new_num) ` `        ``return` `true``; `   `    ``// else all bits are not set ` `    ``return` `false``; ` `} `   `// function to check whether all ` `// the bits in the given range ` `// of two numbers are complement ` `// of each other ` `static` `bool` `bitsAreComplement(``int` `a, ``int` `b, ` `                              ``int` `l, ``int` `r) ` `{ ` `    ``int` `xor_value = a ^ b; ` `    ``return` `allBitsSetInTheGivenRange(xor_value, l, r); ` `} `   `// Driver Code ` `static` `public` `void` `Main ()` `{` `    ``int` `a = 10, b = 5; ` `    ``int` `l = 1, r = 3; ` `    `  `    ``if` `(bitsAreComplement(a, b, l, r)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `} ` `} `   `// This code is contributed ` `// by Ajit Deshpal.`

## PHP

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## Javascript

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Output:

`Yes`

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