# Sum of decimals that are binary representations of first N natural numbers

Given a positive integer N, the task is to calculate the sum of all decimals which can be expressed as binary representations of first N natural numbers.

Examples:

Input: N = 3
Output: 22
Explanation:
The Binary Representation of 1 is 01.
The Binary Representation of 2 is 10.
The Binary Representation of 3 is 11.
Therefore, required sum = 01 + 10 + 11 = 22.

Input: N = 5
Output: 223

Naive Approach: The simplest approach to solve the problem is to iterate a loop over the range [1, N] and in each iteration convert the current number to its binary representation and add it to the overall sum. After adding all the numbers, print the sum as the result.

Time Complexity: O(N * log(N))
Auxiliary Space: O(32)

Efficient Approach: The above approach can also be optimized by finding the contribution of numbers not having the same most significant bit (MSB) position as N and then find the contribution by the MSB of the rest of the numbers. Follow the steps to solve the problem:

• Initialize a variable, say ans as 0 to store the sum of all the numbers in the binary representation of first N natural numbers.
• Iterate until the value of N is at least 0, and perform the following steps:
• Store the MSB position of the number N in a variable X and store the value of 2(X – 1) in a variable, say A.
• Initialize a variable, say cur as 0 to store the contribution of numbers not having the same MSB position as N.
• Iterate over the range [1, X], and in each iteration, add A to the variable cur and then multiply A by 10.
• After the above steps, add the value of cur to ans and store the remaining elements in variable rem as (N – 2X + 1).
• Add the contribution by the MSB of the rest of the numbers by adding (rem * 10X) to the ans.
• Update the value of N to (rem – 1) for the next iteration.
• After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;const int MOD = 1e9 + 7; // Function to find the sum of first// N natural numbers represented// in binary representationvoid sumOfBinaryNumbers(int n){    // Stores the resultant sum    int ans = 0;     int one = 1;     // Iterate until the value of    // N is greater than 0    while (1) {         // If N is less than 2        if (n <= 1) {            ans = (ans + n) % MOD;            break;        }         // Store the MSB position of N        int x = log2(n);         int cur = 0;        int add = (one << (x - 1));         // Iterate in the range [1, x]        // and add the contribution of        // the  numbers from 1 to (2^x-1)        for (int i = 1; i <= x; i++) {             // Update the value of the            // cur and add            cur = (cur + add) % MOD;            add = (add * 10 % MOD);        }         // Add the cur to ans        ans = (ans + cur) % MOD;         // Store the remaining numbers        int rem = n - (one << x) + 1;         // Add the contribution by MSB        // by the remaining numbers        int p = pow(10, x);        p = (p * (rem % MOD)) % MOD;        ans = (ans + p) % MOD;         // The next iteration will        // be repeated for 2^x - 1        n = rem - 1;    }     // Print the result    cout << ans;} // Driver Codeint main(){    int N = 3;    sumOfBinaryNumbers(N);     return 0;}

## Java

 /// Java program for the above approachimport java.io.*;import java.lang.*; class GFG{    static final int MOD = 1000000007;     // Function to find the sum of first    // N natural numbers represented    // in binary representation    static void sumOfBinaryNumbers(int n)    {         // Stores the resultant sum        int ans = 0;         int one = 1;         // Iterate until the value of        // N is greater than 0        while (true) {             // If N is less than 2            if (n <= 1) {                ans = (ans + n) % MOD;                break;            }             // Store the MSB position of N            int x = (int)(Math.log(n) / Math.log(2));             int cur = 0;            int add = (int)(Math.pow(2, (x - 1)));             // Iterate in the range [1, x]            // and add the contribution of            // the  numbers from 1 to (2^x-1)            for (int i = 1; i <= x; i++) {                 // Update the value of the                // cur and add                cur = (cur + add) % MOD;                add = (add * 10 % MOD);            }             // Add the cur to ans            ans = (ans + cur) % MOD;             // Store the remaining numbers            int rem = n - (int)(Math.pow(2, x)) + 1;             // Add the contribution by MSB            // by the remaining numbers            int p = (int)Math.pow(10, x);            p = (p * (rem % MOD)) % MOD;            ans = (ans + p) % MOD;             // The next iteration will            // be repeated for 2^x - 1            n = rem - 1;        }         // Print the result        System.out.println(ans);    }     // Driver Code    public static void main(String[] args)    {        int N = 3;             sumOfBinaryNumbers(N);    }} // This code is contributed by Dharanendra L V

## Python3

 # Python3 program for the above approachfrom math import log2, pow MOD = 1000000007 # Function to find the sum of first# N natural numbers represented# in binary representationdef sumOfBinaryNumbers(n):         # Stores the resultant sum    ans = 0     one = 1     # Iterate until the value of    # N is greater than 0    while (1):                 # If N is less than 2        if (n <= 1):            ans = (ans + n) % MOD            break         # Store the MSB position of N        x = int(log2(n))         cur = 0        add = (one << (x - 1))         # Iterate in the range [1, x]        # and add the contribution of        # the  numbers from 1 to (2^x-1)        for i in range(1, x + 1, 1):                         # Update the value of the            # cur and add            cur = (cur + add) % MOD            add = (add * 10 % MOD)         # Add the cur to ans        ans = (ans + cur) % MOD         # Store the remaining numbers        rem = n - (one << x) + 1         # Add the contribution by MSB        # by the remaining numbers        p = pow(10, x)        p = (p * (rem % MOD)) % MOD        ans = (ans + p) % MOD         # The next iteration will        # be repeated for 2^x - 1        n = rem - 1     # Print the result    print(int(ans)) # Driver Codeif __name__ == '__main__':         N = 3         sumOfBinaryNumbers(N) # This code is contributed by SURENDRA_GANGWAR

## C#

 // C# program for the above approachusing System;class GFG{     const int MOD = 1000000007; // Function to find the sum of first// N natural numbers represented// in binary representationstatic void sumOfBinaryNumbers(int n){         // Stores the resultant sum    int ans = 0;     int one = 1;     // Iterate until the value of    // N is greater than 0    while (true)     {                 // If N is less than 2        if (n <= 1)        {            ans = (ans + n) % MOD;            break;        }         // Store the MSB position of N        int x = (int)Math.Log(n, 2);         int cur = 0;        int add = (one << (x - 1));         // Iterate in the range [1, x]        // and add the contribution of        // the  numbers from 1 to (2^x-1)        for(int i = 1; i <= x; i++)        {                         // Update the value of the            // cur and add            cur = (cur + add) % MOD;            add = (add * 10 % MOD);        }         // Add the cur to ans        ans = (ans + cur) % MOD;         // Store the remaining numbers        int rem = n - (one << x) + 1;         // Add the contribution by MSB        // by the remaining numbers        int p = (int)Math.Pow(10, x);        p = (p * (rem % MOD)) % MOD;        ans = (ans + p) % MOD;         // The next iteration will        // be repeated for 2^x - 1        n = rem - 1;    }     // Print the result    Console.WriteLine(ans);} // Driver Codepublic static void Main(){    int N = 3;         sumOfBinaryNumbers(N);}} // This code is contributed by ukasp

## Javascript



Output:
22

Time Complexity: O(log N)
Auxiliary Space: O(1)

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