Given a number N, the task is to calculate the total number of corresponding different bit in the binary representation for every consecutive number from 0 to N.
Examples:
Input: N = 5
Output: 8
Explanation:
Binary Representation of numbers are:
0 -> 000,
1 -> 001,
2 -> 010,
3 -> 011,
4 -> 100,
5 -> 101
Between 1 and 0 -> 1 bit is different
Between 2 and 1 -> 2 bits are different
Between 3 and 2 -> 1 bit is different
Between 4 and 3 -> 3 bits are different
Between 5 and 4 -> 1 bit is different
Total = 1 + 2 + 1 + 3 + 1 = 8Input: N = 11
Output: 19
For Naive and Efficient Approach please refer to the previous post of this article.
More Efficient Approach: To optimize the above methods, we can use Recursion. To solve the problem, the following observations need to be made
Number: 0 1 2 3 4 5 6 7 Difference: 1 2 1 3 1 2 1 4 Sum: 1 3 4 7 8 10 11 15
We can observe that for N = [1, 2, 3, 4, …..], the sum of different bits in consecutive elements forms the sequence [1, 3, 4, 7, 8, ……]. Hence, Nth term of this series will be our required answer, which can be calculated as:
a(n) = a(n / 2) + n; with base case as a(1) = 1
Below is the implementation for above Recursive approach:
C++
// C++ program to find the sum // of bit differences between // consecutive numbers // from 0 to N using recursion #include <bits/stdc++.h> using namespace std; // Recursive function to find sum // of different bits between // consecutive numbers from 0 to N int totalCountDifference(int n) { // Base case if (n == 1) return 1; // Calculate the Nth term return n + totalCountDifference(n / 2); } // Driver Code int main() { // Given Number int N = 5; // Function Call cout << totalCountDifference(N); return 0; } |
Java
// Java program to find the sum // of bit differences between // consecutive numbers from // 0 to N using recursion class GFG{ // Recursive function to find sum // of different bits between // consecutive numbers from 0 to N static int totalCountDifference(int n) { // Base case if (n == 1) return 1; // Calculate the Nth term return n + totalCountDifference(n / 2); } // Driver Code public static void main(String[] args) { // Given number int N = 5; // Function call System.out.println(totalCountDifference(N)); } } // This code is contributed by himanshu77 |
Python3
# Python3 program to find the sum # of bit differences between # consecutive numbers from # 0 to N using recursion # Recursive function to find sum # of different bits between # consecutive numbers from 0 to N def totalCountDifference (n): # Base case if (n == 1): return 1 # Calculate the Nth term return n + totalCountDifference(n // 2) # Driver code # Given number N = 5 # Function call print(totalCountDifference(N)) # This code is contributed by himanshu77 |
C#
// C# program to find the sum // of bit differences between // consecutive numbers from // 0 to N using recursion using System; class GFG{ // Recursive function to find sum // of different bits between // consecutive numbers from 0 to N static int totalCountDifference(int n) { // Base case if (n == 1) return 1; // Calculate the Nth term return n + totalCountDifference(n / 2); } // Driver Code public static void Main() { // Given number int N = 5; // Function call Console.WriteLine(totalCountDifference(N)); } } // This code is contributed by himanshu77 |
8
Time Complexity: O(log2N)
Auxiliary Space: O(1)
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Improved By : himanshu77
