Count of numbers in range [L, R] with LSB as 0 in their Binary representation
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++
// 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
// 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
# 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#
// 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. |
Javascript
<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++
// 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
// 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
# 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#
// 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. |
Javascript
<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)
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