Given a non-negative number n and two values l and r. The problem is to count the number of unset bits in the range l to r in the binary representation of n, i.e, to count unset bits from the rightmost lth bit to the rightmost rth bit.
Examples:
Input : n = 42, l = 2, r = 5 Output : 2 (42)10 = (101010)2 There are '2' unset bits in the range 2 to 5. Input : n = 80, l = 1, r = 4 Output : 4
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.
- Count number of set bits in the number (n & num). Refer this post. Let it be count.
- Calculate ans = (r – l + 1) – count.
- Return ans.
C++
// C++ implementation to count unset bits in the // given range #include <bits/stdc++.h> using namespace std;
// Function to get no of set bits in the // binary representation of 'n' unsigned int countSetBits( int n)
{ unsigned int count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
} // function to count unset bits // in the given range unsigned int countUnsetBitsInGivenRange(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);
// returns number of unset bits in the range
// 'l' to 'r' in 'n'
return (r - l + 1) - countSetBits(n & num);
} // Driver program to test above int main()
{ unsigned int n = 80;
unsigned int l = 1, r = 4;
cout << countUnsetBitsInGivenRange(n, l, r);
return 0;
} |
Java
// Java implementation to count unset bits in the // given range class GFG {
// Function to get no of set bits in the
// binary representation of 'n'
static int countSetBits( int n)
{
int count = 0 ;
while (n > 0 ) {
n &= (n - 1 );
count++;
}
return count;
}
// function to count unset bits
// in the given range
static int countUnsetBitsInGivenRange( 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 );
// returns number of unset bits in the range
// 'l' to 'r' in 'n'
return (r - l + 1 ) - countSetBits(n & num);
}
// Driver code
public static void main(String[] args)
{
int n = 80 ;
int l = 1 , r = 4 ;
System.out.print(
countUnsetBitsInGivenRange(n, l, r));
}
} // This code is contributed by Anant Agarwal. |
Python3
# Python3 implementation to count # unset bits in the given range # Function to get no of set bits in # the binary representation of 'n' def countSetBits (n):
count = 0
while n:
n & = (n - 1 )
count + = 1
return count
# function to count unset bits # in the given range def countUnsetBitsInGivenRange (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 )
# returns number of unset bits
# in the range 'l' to 'r' in 'n'
return (r - l + 1 ) - countSetBits(n & num)
# Driver code to test above n = 80
l = 1
r = 4
print (countUnsetBitsInGivenRange(n, l, r))
# This code is contributed by "Sharad_Bhardwaj" |
C#
// C# implementation to count unset bits in the // given range using System;
class GFG {
// Function to get no of set bits in the
// binary representation of 'n'
static int countSetBits( int n)
{
int count = 0;
while (n > 0) {
n &= (n - 1);
count++;
}
return count;
}
// function to count unset bits
// in the given range
static int countUnsetBitsInGivenRange( 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);
// returns number of unset bits in the range
// 'l' to 'r' in 'n'
return (r - l + 1) - countSetBits(n & num);
}
//Driver code
public static void Main()
{
int n = 80;
int l = 1, r = 4;
Console.Write(countUnsetBitsInGivenRange(n, l, r));
}
} //This code is contributed by Anant Agarwal. |
PHP
<?php // php implementation to count // unset bits in the given range // Function to get no of set bits in // the binary representation of 'n' function countSetBits( $n )
{ $count = 0;
while ( $n )
{
$n &= ( $n - 1);
$count ++;
}
return $count ;
} // function to count unset // bits in the given range function countUnsetBitsInGivenRange( $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);
// returns number of unset
// bits in the range
// 'l' to 'r' in 'n'
return ( $r - $l + 1) -
countSetBits( $n & $num );
} // Driver code
$n = 80;
$l = 1;
$r = 4;
echo countUnsetBitsInGivenRange( $n , $l , $r );
// This code is contributed by mits ?> |
Javascript
<script> // Javascript implementation to count unset bits in the // given range // Function to get no of set bits in the // binary representation of 'n' function countSetBits(n)
{ var count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
} // function to count unset bits // in the given range function countUnsetBitsInGivenRange(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);
// returns number of unset bits in the range
// 'l' to 'r' in 'n'
return (r - l + 1) - countSetBits(n & num);
} // Driver program to test above var n = 80;
var l = 1, r = 4;
document.write( countUnsetBitsInGivenRange(n, l, r)); </script> |
Output:
4
Time Complexity: O(log n)
Space Complexity: O(1)
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