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Find the number obtained by concatenating binary representations of all numbers up to N
  • Last Updated : 12 Jan, 2021

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 corresponds 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 ith 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 ith 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, left shift ans by l and then bitwise OR the result with i.
  5. After traversing, print ans as the answer.

Below is the implementation of the above approach:

C++

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// 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

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output: 

27

 

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


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