Given two integers L and R. The task is to find the count of all numbers in the range [L, R] whose Least Significant Bit in binary representation is 0.
Examples:
Input: L = 10, R = 20
Output: 6Input: L = 7, R = 11
Output: 2
Naive approach: The simplest approach is to solve this problem is to check for every number in the range [L, R], if Least Significant Bit in binary representation is 0.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the count // of required numbers int countNumbers( int l, int r)
{ int count = 0;
for ( int i = l; i <= r; i++) {
// If rightmost bit is 0
if ((i & 1) == 0) {
count++;
}
}
// Return the required count
return count;
} // Driver code int main()
{ int l = 10, r = 20;
// Call function countNumbers
cout << countNumbers(l, r);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG{
// Function to return the count // of required numbers static int countNumbers( int l, int r)
{ int count = 0 ;
for ( int i = l; i <= r; i++)
{
// If rightmost bit is 0
if ((i & 1 ) == 0 )
count += 1 ;
}
// Return the required count
return count;
} // Driver code public static void main(String[] args)
{ int l = 10 , r = 20 ;
// Call function countNumbers
System.out.println(countNumbers(l, r));
} } // This code is contributed by MuskanKalra1 |
# Python3 implementation of the approach # Function to return the count # of required numbers def countNumbers(l, r):
count = 0
for i in range (l, r + 1 ):
# If rightmost bit is 0
if ((i & 1 ) = = 0 ):
count + = 1
# Return the required count
return count
# Driver code l = 10
r = 20
# Call function countNumbers print (countNumbers(l, r))
# This code is contributed by amreshkumar3 |
// C# implementation of the approach using System;
class GFG {
// Function to return the count
// of required numbers
static int countNumbers( int l, int r)
{
int count = 0;
for ( int i = l; i <= r; i++) {
// If rightmost bit is 0
if ((i & 1) == 0)
count += 1;
}
// Return the required count
return count;
}
// Driver code
public static void Main()
{
int l = 10, r = 20;
// Call function countNumbers
Console.WriteLine(countNumbers(l, r));
}
} // This code is contributed by subham348. |
<script> // JavaScript implementation of the approach // Function to return the count // of required numbers function countNumbers(l, r)
{ let count = 0;
for (let i = l; i <= r; i++) {
// If rightmost bit is 0
if ((i & 1) == 0) {
count++;
}
}
// Return the required count
return count;
} // Driver code let l = 10, r = 20;
// Call function countNumbers
document.write(countNumbers(l, r));
</script> |
6
Time Complexity: O(r – l)
Auxiliary Space: O(1)
Efficient approach: This problem can be solved by using properties of bits. Only even numbers have rightmost bit as 0. The count can be found using this formula ((R / 2) – (L – 1) / 2) in O(1) time.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the count // of required numbers int countNumbers( int l, int r)
{ // Count of numbers in range
// which are divisible by 2
return ((r / 2) - (l - 1) / 2);
} // Driver code int main()
{ int l = 10, r = 20;
cout << countNumbers(l, r);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG{
// Function to return the count // of required numbers static int countNumbers( int l, int r)
{ // Count of numbers in range
// which are divisible by 2
return ((r / 2 ) - (l - 1 ) / 2 );
} // Driver Code public static void main(String[] args)
{ int l = 10 ;
int r = 20 ;
System.out.println(countNumbers(l, r));
} } // This code is contributed by MuskanKalra1 |
# Python3 implementation of the approach # Function to return the count # of required numbers def countNumbers(l, r):
# Count of numbers in range
# which are divisible by 2
return ((r / / 2 ) - (l - 1 ) / / 2 )
# Driver code l = 10
r = 20
print (countNumbers(l, r))
# This code is contributed by amreshkumar3 |
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG{
// Function to return the count // of required numbers static int countNumbers( int l, int r)
{ // Count of numbers in range
// which are divisible by 2
return ((r / 2) - (l - 1) / 2);
} // Driver code public static void Main()
{ int l = 10, r = 20;
Console.Write(countNumbers(l, r));
} } // This code is contributed by SURENDRA_GANGWAR. |
<script> // Javascript implementation of the approach // Function to return the count // of required numbers function countNumbers(l, r)
{ // Count of numbers in range
// which are divisible by 2
return (parseInt(r / 2) -
parseInt((l - 1) / 2));
} // Driver code let l = 10, r = 20; document.write(countNumbers(l, r)); // This code is contributed by subhammahato348 </script> |
6
Time Complexity: O(1)
Auxiliary Space: O(1)