Find a number containing N – 1 set bits at even positions from the right

Given a positive integer N, the task is to find a number which contains (N – 1) set bits in its binary form at every even index (1-based) from the right.

Examples:

Input: N = 2
Output: 2
Binary representation of 2 is 10 which has
1 set bit at even position from the right.



Input: N = 4
Output: 42
Binary representation of 42 is 101010

Observation: If we check out the numbers in binary form then the result is something like this:

n Decimal Equivalent Binary Equivalent
1 0 0
2 2 10
3 10 1010
4 42 101010
5 170 10101010

Naive Approach: As we can see in the table our binary equivalent is always adding a “10” in last of the previous string. So, we can generate a binary string which is made up of sub-string “10” concatenated N-1 times and then print its decimal equivalent.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define ll long long int
  
// Function to return the string generated
// by appending "10" n-1 times
string constructString(ll n)
{
    // Initialising string as empty
    string s = "";
    for (ll i = 0; i < n; i++) {
        s += "10";
    }
    return s;
}
  
// Function to return the decimal equivalent
// of the given binary string
ll binaryToDecimal(string n)
{
    string num = n;
    ll dec_value = 0;
  
    // Initializing base value to 1
    // i.e 2^0
    ll base = 1;
  
    ll len = num.length();
    for (ll i = len - 1; i >= 0; i--) {
        if (num[i] == '1')
            dec_value += base;
        base = base * 2;
    }
  
    return dec_value;
}
  
// Function that calls the constructString
// and binarytodecimal and returns the answer
ll findNumber(ll n)
{
    string s = constructString(n - 1);
    ll num = binaryToDecimal(s);
    return num;
}
  
// Driver code
int main()
{
    ll n = 4;
  
    cout << findNumber(n);
  
    return 0;
}

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Java

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// Java implementation of above approach
import java.util.*;
  
class GFG 
{
  
// Function to return the String generated
// by appending "10" n-1 times
static String constructString(int n)
{
    // Initialising String as empty
    String s = "";
    for (int i = 0; i < n; i++) 
    {
        s += "10";
    }
    return s;
}
  
// Function to return the decimal equivalent
// of the given binary String
static int binaryToDecimal(String n)
{
    String num = n;
    int dec_value = 0;
  
    // Initializing base value to 1
    // i.e 2^0
    int base = 1;
  
    int len = num.length();
    for (int i = len - 1; i >= 0; i--) 
    {
        if (num.charAt(i) == '1')
            dec_value += base;
        base = base * 2;
    }
  
    return dec_value;
}
  
// Function that calls the constructString
// and binarytodecimal and returns the answer
static int findNumber(int n)
{
    String s = constructString(n - 1);
    int num = binaryToDecimal(s);
    return num;
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 4;
  
    System.out.println(findNumber(n));
}
}
  
/* This code is contributed by PrinciRaj1992 */

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Python

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# Python3 implementation of the approach
  
# Function to return the generated
# by appending "10" n-1 times
def constructString(n):
  
    # Initialising as empty
    s = ""
    for i in range(n):
        s += "10"
  
    return s
  
# Function to return the decimal equivaLent
# of the given binary string
def binaryToDecimal(n):
  
    num = n
    dec_value = 0
  
    # Initializing base value to 1
    # i.e 2^0
    base = 1
  
    Len = len(num)
    for i in range(Len - 1,-1,-1):
        if (num[i] == '1'):
            dec_value += base
        base = base * 2
  
  
    return dec_value
  
# Function that calls the constructString
# and binarytodecimal and returns the answer
def findNumber(n):
  
    s = constructString(n - 1)
    num = binaryToDecimal(s)
    return num
  
# Driver code
n = 4
  
print(findNumber(n))
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of above approach
using System;
  
class GFG
{
      
// Function to return the String generated
// by appending "10" n-1 times
static String constructString(int n)
{
      
    // Initialising String as empty
    String s = "";
    for (int i = 0; i < n; i++) 
    {
        s += "10";
    }
    return s;
}
  
// Function to return the decimal equivalent
// of the given binary String
static int binaryToDecimal(String n)
{
    String num = n;
    int dec_value = 0;
  
    // Initializing base value to 1
    // i.e 2^0
    int base_t = 1;
  
    int len = num.Length;
    for (int i = len - 1; i >= 0; i--) 
    {
        if (num[i] == '1')
            dec_value = dec_value + base_t;
        base_t = base_t * 2;
    }
  
    return dec_value;
}
  
// Function that calls the constructString
// and binarytodecimal and returns the answer
static int findNumber(int n)
{
    String s = constructString(n - 1);
    int num = binaryToDecimal(s);
    return num;
}
  
// Driver code
static public void Main ()
{
    int n = 4;
    Console.Write(findNumber(n));
}
}
  
// This code is contributed by ajit

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Output:

42

Efficient Approach: If we take the numbers and convert them to base 4 we can see an interesting pattern as follows:

n Decimal Equivalent Binary Equivalent Base_4
1 0 0 0
2 2 10 2
3 10 1010 22
4 42 101010 222
5 170 10101010 2222

We are actually appending “2” for every nth term in base4 i.e. for n = 7 our number in base4 would have (n – 1) i.e. 6 consecutive 2’s.
Now we have to take a point in mind as we know that if we convert from any base m to base 10 i.e. decimal than the solution is (n0 * m0 + n1 * m1 + n2 * m2 + …. + n * mn). So as our base is 4 by further calcuation we can found that our required number n can be found by using the deduced formula in O(1) time complexity.
Formula:

A(n) = floor((2 / 3) * (4n – 1))

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define ll long long int
  
// Function to compute number
// using our deduced formula
ll findNumber(int n)
{
    // Initialize num to n-1
    ll num = n - 1;
    num = 2 * (ll)pow(4, num);
    num = floor(num / 3.0);
    return num;
}
  
// Driver code
int main()
{
    int n = 5;
    cout << findNumber(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to compute number
// using our deduced formula
static int findNumber(int n)
{
    // Initialize num to n-1
    int num = n - 1;
    num = 2 * (int)Math.pow(4, num);
    num = (int)Math.floor(num / 3.0);
    return num;
}
  
// Driver code
public static void main (String[] args) 
{
    int n = 5;
    System.out.println (findNumber(n));
}
}
  
// The code is contributed by ajit.

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Python3

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# Python3 implementation of the approach 
  
# Function to compute number 
# using our deduced formula 
def findNumber(n) :
      
    # Initialize num to n-1 
    num = n - 1;
    num = 2 * (4 ** num);
    num = num // 3;
    return num; 
  
# Driver code 
if __name__ == "__main__" :
      
    n = 5;
    print(findNumber(n)); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to compute number
// using our deduced formula
static int findNumber(int n)
{
    // Initialize num to n-1
    int num = n - 1;
    num = 2 * (int)Math.Pow(4, num);
    num = (int)Math.Floor(num / 3.0);
    return num;
}
  
// Driver code
static public void Main ()
{
          
    int n = 5;
    Console.Write(findNumber(n));
}
}
  
// The code is contributed by Tushil.

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Output:

170


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