Given a non-negative number n and two values l and r. The problem is to check whether all the bits are unset or not in the range l to r in the binary representation of n.
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.
Examples:
Input : n = 17, l = 2, r = 4 Output : Yes (17)10 = (10001)2 The bits in the range 2 to 4 are all unset. Input : n = 36, l = 3, r = 5 Output : No (36)10 = (100100)2 The bits in the range 3 to 5 are all not unset.
Approach: Following are the steps:
- Calculate num = ((1 << r) – 1) ^ ((1 << (l-1)) – 1). This will produce a number num having r number of bits and bits in the range l to r are the only set bits.
- Calculate new_num = n & num.
- If new_num == 0, return “Yes” (all bits are unset in the given range).
- Else return “No” (all bits are not unset in the given range).
C++
// C++ implementation to check whether all // the bits are unset in the given range // or not #include <bits/stdc++.h> using namespace std;
// function to check whether all the bits // are unset in the given range or not string 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 new num is 0, then all bits
// are unset in the given range
if (new_num == 0)
return "Yes" ;
// else all bits are not unset
return "No" ;
} // Driver program to test above int main()
{ unsigned int n = 17;
unsigned int l = 2, r = 4;
cout << allBitsSetInTheGivenRange(n, l, r);
return 0;
} |
Java
// Java implementation to // check whether all the // bits are unset in the // given range or not import java.io.*;
class GFG
{ // function to check whether // all the bits are unset in // the given range or not static String 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 new num is 0, then
// all bits are unset in
// the given range
if (new_num == 0 )
return "Yes" ;
// else all bits
// are not unset
return "No" ;
} // Driver Code public static void main (String[] args)
{ int n = 17 ;
int l = 2 ;
int r = 4 ;
System.out.println(
allBitsSetInTheGivenRange(n, l, r));
} } // This code is contributed by akt_mit |
Python 3
# Python 3 implementation to check whether # all the bits are unset in the given range # or not # function to check whether all the bits # are unset 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 new num is 0, then all bits
# are unset in the given range
if (new_num = = 0 ):
return "Yes"
# else all bits are not unset
return "No"
# Driver Code if __name__ = = "__main__" :
n = 17
l = 2
r = 4
print (allBitsSetInTheGivenRange(n, l, r))
# This code is contributed by ita_c |
C#
// C# implementation to // check whether all the // bits are unset in the // given range or not using System;
public class GFG{
// function to check whether // all the bits are unset in // the given range or not static String 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 new num is 0, then
// all bits are unset in
// the given range
if (new_num == 0)
return "Yes" ;
// else all bits
// are not unset
return "No" ;
} // Driver Code static public void Main (){
int n = 17;
int l = 2;
int r = 4;
Console.WriteLine(
allBitsSetInTheGivenRange(n, l, r));
}
} // This code is contributed by k_mit |
PHP
<?php // PHP implementation to check // whether all the bits are // unset in the given range // or not // function to check whether // all the bits are unset 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 new num is 0, then
// all bits are unset in
// the given range
if ( $new_num == 0)
return "Yes" ;
// else all bits
// are not unset
return "No" ;
} // Driver Code $n = 17;
$l = 2; $r = 4;
echo allBitsSetInTheGivenRange( $n ,
$l , $r );
// This code is contributed // by ajit ?> |
Javascript
<script> // Javascript implementation to check whether all // the bits are unset in the given range // or not // function to check whether all the bits // are unset 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
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 new num is 0, then all bits
// are unset in the given range
if (new_num == 0)
return "Yes" ;
// else all bits are not unset
return "No" ;
} // Driver program to test above var n = 17;
var l = 2, r = 4;
document.write( allBitsSetInTheGivenRange(n, l, r)); </script> |
Output:
Yes
Time Complexity : O(1)
Auxiliary Space: O(1)
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