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

• Last Updated : 18 Aug, 2021

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:
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:

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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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 */`

## Python

 `# 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`

## C#

 `// 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`

## Javascript

 ``
Output:
`42`

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

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 calculation 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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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.`

## Python3

 `# 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`

## C#

 `// 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.`

## Javascript

 ``
Output:
`170`

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