Find the number obtained by concatenating binary representations of all numbers up to N

Given an integer N, the task is to find the decimal value of the binary string formed by concatenating the binary representations of all numbers from 1 to N sequentially.

Examples:

Input: N = 12
Output: 118505380540
Explanation: The concatenation results in "1101110010111011110001001101010111100". The equivalent decimal value is 118505380540.

Input: N = 3
Output: 27
Explanation: In binary, 1, 2, and 3 correspond to "1", "10", and "11". Their concatenation results in "11011", which corresponds to the decimal value of 27.

Approach: The idea is to iterate over the range [1, N]. For every i^th number, concatenate the binary representation of the number i using the Bitwise XOR property. Follow the steps below to solve the problem:

1. Initialize two variables, l, and ans with 0, where l stores the current position of the bit in the final binary string of any i^th number and ans will store the final answer.
2. Iterate from i = 1 to N + 1.
3. If (i & ( i - 1 )) is equal to 0, then simply increment the value of l by 1, where & is the Bitwise AND operator.
4. After that, the left shift ans by l and then bitwise OR the result with i.
5. After and traversing, print ans as the answer.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the decimal value by
// concatenating the numbers from 1 to N
int concatenatedBinary(int n)
{

    // Stores count of
    // bits in a number
    int l = 0;

    // Stores decimal value by
    // concatenating 1 to N
    int ans = 0;

    // Iterate over the range [1, n]
    for (int i = 1; i < n + 1; i++){

        // If i is a power of 2
        if ((i & (i - 1)) == 0)
              l += 1;

        // Update ans
        ans = ((ans << l) | i);
    }

    // Return ans
    return ans;
}

// Driver Code
int main()
{
    int n = 3;

    // Function Call
    cout << (concatenatedBinary(n));

  return 0;
}

// This code is contributed by mohiy kumar 29

// Java program for the above approach
class GFG
{
    
    // Function to find the decimal value by
    // concatenating the numbers from 1 to N
    static int concatenatedBinary(int n)
    {
    
        // Stores count of
        // bits in a number
        int l = 0;
    
        // Stores decimal value by
        // concatenating 1 to N
        int ans = 0;
    
        // Iterate over the range [1, n]
        for (int i = 1; i < n + 1; i++){
    
            // If i is a power of 2
            if ((i & (i - 1)) == 0)
                  l += 1;
    
            // Update ans
            ans = ((ans << l) | i);
        }
    
        // Return ans
        return ans;
    }
    
    // Driver Code
    public static void main (String[] args)
    {
        int n = 3;
    
        // Function Call
        System.out.println(concatenatedBinary(n));
    }
}

// This code is contributed by AnkThon

# Python program for the above approach

# Function to find the decimal value by 
# concatenating the numbers from 1 to N 
def concatenatedBinary(n):

    # Stores count of 
    # bits in a number
    l = 0

    # Stores decimal value by
    # concatenating 1 to N 
    ans = 0

    # Iterate over the range [1, n]
    for i in range(1, n + 1): 

        # If i is a power of 2
        if i & (i - 1) == 0: 

            # Update l
            l += 1

        # Update ans
        ans = ((ans << l) | i) 

    # Return ans
    return(ans) 

# Driver Code
if __name__ == '__main__':
    n = 3

    # Function Call
    print(concatenatedBinary(n))

// C# program to implement
// the above approach  
using System;
class GFG
{
    
    // Function to find the decimal value by
    // concatenating the numbers from 1 to N
    static int concatenatedBinary(int n)
    {
    
        // Stores count of
        // bits in a number
        int l = 0;
    
        // Stores decimal value by
        // concatenating 1 to N
        int ans = 0;
    
        // Iterate over the range [1, n]
        for (int i = 1; i < n + 1; i++)
        {
    
            // If i is a power of 2
            if ((i & (i - 1)) == 0)
                  l += 1;
    
            // Update ans
            ans = ((ans << l) | i);
        }
    
        // Return ans
        return ans;
    }
    
    // Driver Code
    public static void Main ()
    {
        int n = 3;
    
        // Function Call
        Console.WriteLine(concatenatedBinary(n));
    }
}

// This code is contributed by sanjoy_62

<script>
//Javascript program to implement
// the above approach

// Function to find the decimal value by
// concatenating the numbers from 1 to N
function concatenatedBinary(n)
{

    // Stores count of
    // bits in a number
    var l = 0;

    // Stores decimal value by
    // concatenating 1 to N
    var ans = 0;

    // Iterate over the range [1, n]
    for (var i = 1; i < n + 1; i++){

        // If i is a power of 2
        if ((i & (i - 1)) == 0)
              l += 1;

        // Update ans
        ans = parseInt((ans << l) | i);
    }

    // Return ans
    return ans;
}


var n = 3;
// Function Call
document.write(concatenatedBinary(n));

//This code is contributed by SoumikMondal
</script>

27

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