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 <bits/stdc++.h>using namespace std;// function to check whether all the bits// are set in the given range or notbool 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 otherbool 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 Codeint 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 otherclass GFG{// function to check whether // all the bits are set in // the given range or notstatic 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 otherstatic boolean bitsAreComplement(int a, int b, int l, int r){ int xor_value = a ^ b; return allBitsSetInTheGivenRange(xor_value, l, r);}// Driver Codepublic 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 notdef 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 otherdef bitsAreComplement(a, b, l, r): xor_value = a ^ b return allBitsSetInTheGivenRange(xor_value, l, r)# Driver Codeif __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
<?php// PHP 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 function 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 == $new_num) return true; // else all bits // are not set return false; } // function to check whether// all the bits in the given range // of two numbers are complement // of each other function bitsAreComplement($a, $b,$l, $r) { $xor_value = $a ^ $b; return allBitsSetInTheGivenRange($xor_value, $l, $r); } // Driver Code $a = 10;$b = 5; $l = 1;$r = 3; if (bitsAreComplement($a, $b, $l, $r)) echo "Yes"; else echo "No"; // This Code is Contributed by ajit?> |
Javascript
<script>// javascript 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 notfunction 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 var 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 var 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 otherfunction bitsAreComplement(a , b,l , r){ var xor_value = a ^ b; return allBitsSetInTheGivenRange(xor_value, l, r);}// Driver Codevar a = 10, b = 5;var l = 1, r = 3;if (bitsAreComplement(a, b, l, r)) document.write("Yes");else document.write("No");// This code is contributed by Princi Singh </script> |
Output:
Yes
Time complexity: O(1) as it is performing constant operations
Auxiliary space: O(1)
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