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.
Try It Yourself
// 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 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 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# 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 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
?>
<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)