Count set bits in a range
Last Updated : 15 Jul, 2022
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Given a non-negative number n and two values l and r. The problem is to count the number of set bits in the range l to r in the binary representation of n, i.e, to count set bits from the rightmost lth bit to the rightmost rth bit.
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.
Examples:
Input : n = 42, l = 2, r = 5 Output : 2 (42)10 = (101010)2 There are '2' set bits in the range 2 to 5. Input : n = 79, 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.
// C++ implementation to count set 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 set bits in the given range
unsigned int countSetBitsInGivenRange(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 set bits in the range
// 'l' to 'r' in 'n'
return countSetBits(n & num);
}
// Driver program to test above
int main()
{
unsigned int n = 42;
unsigned int l = 2, r = 5;
cout << countSetBitsInGivenRange(n, l, r);
return 0;
}
// Java implementation to count set 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 set bits in the given range
static int countSetBitsInGivenRange(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 set bits in the range
// 'l' to 'r' in 'n'
return countSetBits(n & num);
}
// Driver code
public static void main(String[] args)
{
int n = 42;
int l = 2, r = 5;
System.out.print(countSetBitsInGivenRange(n, l, r));
}
}
// This code is contributed by Anant Agarwal.
# Python3 implementation to count
# set 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 = count + 1
return count
# function to count set bits in
# the given range
def countSetBitsInGivenRange(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 set bits in the range
# 'l' to 'r' in 'n'
return countSetBits(n & num)
# Driver program to test above
n = 42
l = 2
r = 5
ans = countSetBitsInGivenRange(n, l, r)
print (ans)
# This code is contributed by Saloni Gupta.
// C# implementation to count set 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 set bits in the given range
static int countSetBitsInGivenRange(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 set bits in the range
// 'l' to 'r' in 'n'
return countSetBits(n & num);
}
//Driver code
public static void Main()
{
int n = 42;
int l = 2, r = 5;
Console.WriteLine(countSetBitsInGivenRange(n, l, r));
}
}
// This code is contributed by Anant Agarwal.
<?php
// PHP implementation to count
// set 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 $count;
}
// function to count set
// bits in the given range
function countSetBitsInGivenRange($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
// set bits in the range
// 'l' to 'r' in 'n'
return countSetBits($n & $num);
}
// Driver Code
$n = 42;
$l = 2;
$r = 5;
echo countSetBitsInGivenRange($n, $l, $r);
// This code is contributed by Ajit.
?>
<script>
// Javascript implementation to
// count set bits in the
// given range
// Function to get no of set bits in the
// binary representation of 'n'
function countSetBits(n)
{
let count = 0;
while (n > 0) {
n &= (n - 1);
count++;
}
return count;
}
// function to count set
// bits in the given range
function countSetBitsInGivenRange(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
let num = ((1 << r) - 1) ^
((1 << (l - 1)) - 1);
// returns number of set
// bits in the range
// 'l' to 'r' in 'n'
return countSetBits(n & num);
}
// driver program
let n = 42;
let l = 2, r = 5;
document.write(countSetBitsInGivenRange(n, l, r));
</script>
Output:
2
Time Complexity: O(logn)
Auxiliary Space: O(1)
