Given a number n. Find the sum of all numbers up to n whose 2 bits are set.
Examples:
Input : 10 Output : 33 3 + 5 + 6 + 9 + 10 = 33 Input : 100 Output : 762
Naive Approach: Find each number up to n whose 2 bits are set. If its 2 bits are set add it to the sum.
C++
// CPP program to find sum of numbers // upto n whose 2 bits are set #include <bits/stdc++.h> using namespace std;
// To count number of set bits int countSetBits( int n)
{ int count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
} // To calculate sum of numbers int findSum( int n)
{ int sum = 0;
// To count sum of number
// whose 2 bit are set
for ( int i = 1; i <= n; i++)
if (countSetBits(i) == 2)
sum += i;
return sum;
} // Driver program to test above function int main()
{ int n = 10;
cout << findSum(n);
return 0;
} |
Java
// Java program to find sum of numbers // upto n whose 2 bits are set public class Main {
// To count number of set bits
static int countSetBits( int n)
{
int count = 0 ;
while (n > 0 ) {
n &= (n - 1 );
count++;
}
return count;
}
// To calculate sum of numbers
static int findSum( int n)
{
int sum = 0 ;
// To count sum of number
// whose 2 bit are set
for ( int i = 1 ; i <= n; i++)
if (countSetBits(i) == 2 )
sum += i;
return sum;
}
// Driver program to test above function
public static void main(String[] args)
{
int n = 10 ;
System.out.println(findSum(n));
}
} |
Python3
# Python program to find # sum of numbers # upto n whose 2 bits are set # To count number of set bits def countSetBits(n):
count = 0
while (n):
n = n & (n - 1 )
count = count + 1
return count
# To calculate sum of numbers def findSum(n):
sum = 0
# To count sum of number
# whose 2 bit are set
for i in range ( 1 ,n + 1 ):
if (countSetBits(i) = = 2 ):
sum = sum + i
return sum
# Driver code n = 10
print (findSum(n))
# This code is contributed # by Anant Agarwal. |
C#
// C# program to find sum of // numbers upto n whose 2 // bits are set using System;
class GFG
{ // To count number
// of set bits
static int countSetBits( int n)
{
int count = 0;
while (n > 0)
{
n = n & (n - 1);
count++;
}
return count;
}
// To calculate
// sum of numbers
static int findSum( int n)
{
int sum = 0;
// To count sum of number
// whose 2 bit are set
for ( int i = 1; i <= n; i++)
if (countSetBits(i) == 2)
sum += i;
return sum;
}
// Driver Code
static public void Main ()
{
int n = 10;
Console.WriteLine(findSum(n));
}
} // This code is contributed by aj_36 |
PHP
<?php // PHP program to find sum of numbers // upto n whose 2 bits are set // To count number of set bits function countSetBits( $n )
{ $count = 0;
while ( $n )
{
$n &= ( $n - 1);
$count ++;
}
return $count ;
} // To calculate sum of numbers function findSum( $n )
{ $sum = 0;
// To count sum of number
// whose 2 bit are set
for ( $i = 1; $i <= $n ; $i ++)
if (countSetBits( $i ) == 2)
$sum += $i ;
return $sum ;
} // Driver Code
$n = 10;
echo findSum( $n );
// This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript program to find sum of numbers // upto n whose 2 bits are set // To count number of set bits function countSetBits(n)
{ let count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
} // To calculate sum of numbers function findSum(n)
{ let sum = 0;
// To count sum of number
// whose 2 bit are set
for (let i = 1; i <= n; i++)
if (countSetBits(i) == 2)
sum += i;
return sum;
} // Driver program to test above function let n = 10;
document.write(findSum(n));
// This code is contributed by Mayank Tyagi </script> |
Output:
33
Time Complexity : O(n)
Space Complexity : O(1)
Efficient Approach: The number whose 2 bits are set is of the form 2^x + 2^y and this number is less than n. So we have to find only numbers in the range up to n which is of form 2^i + 2^j where i > 0 and 2^i < n and 0 <= j < i.
C++
// C++ program to find sum of numbers // upto n whose 2 bits are set #include <bits/stdc++.h> using namespace std;
// To calculate sum of numbers int findSum( int n)
{ int sum = 0;
// Find numbers whose 2 bits are set
for ( int i = 1; (1 << i) < n; i++) {
for ( int j = 0; j < i; j++) {
int num = (1 << i) + (1 << j);
// If number is greater than n
// we don't include this in sum
if (num <= n)
sum += num;
}
}
// Return sum of numbers
return sum;
} // Driver program to test findSum() int main()
{ int n = 10;
cout << findSum(n);
return 0;
} |
Java
// Java program to find sum of numbers // upto n whose 2 bits are set public class Main {
// To calculate sum of numbers
static int findSum( int n)
{
int sum = 0 ;
// Find numbers whose 2 bits are set
for ( int i = 1 ; 1 << i < n; i++) {
for ( int j = 0 ; j < i; j++) {
int num = ( 1 << i) + ( 1 << j);
// If number is greater than n
// we don't include this in sum
if (num <= n)
sum += num;
}
}
// Return sum of numbers
return sum;
}
// Driver program to test findSum()
public static void main(String[] args)
{
int n = 10 ;
System.out.println(findSum(n));
}
} |
Python3
# Python3 program to find sum of # numbers upto n whose 2 bits are set # To calculate sum of numbers def findSum(n) :
sum = 0
# Find numbers whose 2
# bits are set
i = 1
while (( 1 << i) < n ) :
for j in range ( 0 , i) :
num = ( 1 << i) + ( 1 << j)
# If number is greater than n
# we don't include this in sum
if (num < = n) :
sum + = num
i + = 1
# Return sum of numbers
return sum
# Driver Code n = 10
print (findSum(n))
# This code is contributed # by Smitha |
C#
// C# program to find sum of numbers // upto n whose 2 bits are set using System;
public class main {
// To calculate sum of numbers
static int findSum( int n)
{
int sum = 0;
// Find numbers whose 2 bits are set
for ( int i = 1; 1 << i < n; i++)
{
for ( int j = 0; j < i; j++)
{
int num = (1 << i) + (1 << j);
// If number is greater than n
// we don't include this in sum
if (num <= n)
sum += num;
}
}
// Return sum of numbers
return sum;
}
// Driver Code
public static void Main(String []args)
{
int n = 10;
Console.WriteLine(findSum(n));
}
} // This Code is contributed by vt_m. |
PHP
<?php <?php // PHP program to find sum of numbers // upto n whose 2 bits are set // To calculate sum of numbers function findSum( $n )
{ $sum = 0;
// Find numbers whose 2 bits are set
for ( $i = 1; (1 << $i ) < $n ; $i ++)
{
for ( $j = 0; $j < $i ; $j ++)
{
$num = (1 << $i ) + (1 << $j );
// If number is greater than n
// we don't include this in sum
if ( $num <= $n )
$sum += $num ;
}
}
// Return sum of numbers
return $sum ;
} // Driver Code $n = 10;
echo findSum( $n );
// This code is contributed by Ajit ?> |
Javascript
<script> // Javascript program to find sum of numbers
// upto n whose 2 bits are set
// To calculate sum of numbers
function findSum(n)
{
let sum = 0;
// Find numbers whose 2 bits are set
for (let i = 1; 1 << i < n; i++)
{
for (let j = 0; j < i; j++)
{
let num = (1 << i) + (1 << j);
// If number is greater than n
// we don't include this in sum
if (num <= n)
sum += num;
}
}
// Return sum of numbers
return sum;
}
let n = 10;
document.write(findSum(n));
</script> |
Output :
33
Time Complexity : O((log n)*(log n))
Space Complexity : O(1)
Recommended Articles