# Check whether the binary equivalent of a number ends with “001” or not

• Last Updated : 19 Jul, 2022

Given a positive integer N, the task is to check whether the binary equivalent of that integer ends with “001” or not.
Print “Yes” if it ends in “001”. Otherwise, Print “No“.
Examples :

Input: N = 9
Output: Yes
Explanation
Binary of 9 = 1001, which ends with 001
Input: N = 5
Output: No
Binary of 5 = 101, which does not end in 001

Naive Approach
Find the Binary Equivalent of N and check if 001 is a Suffix of its Binary Equivalent.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the``// above approach` `#include ``using` `namespace` `std;` `// Function returns true if``// s1 is suffix of s2``bool` `isSuffix(string s1,``              ``string s2)``{``    ``int` `n1 = s1.length();``    ``int` `n2 = s2.length();``    ``if` `(n1 > n2)``        ``return` `false``;``    ``for` `(``int` `i = 0; i < n1; i++)``        ``if` `(s1[n1 - i - 1]``            ``!= s2[n2 - i - 1])``            ``return` `false``;``    ``return` `true``;``}` `// Function to check if binary equivalent``// of a number ends in "001" or not``bool` `CheckBinaryEquivalent(``int` `N)``{` `    ``// To store the binary``    ``// number``    ``int` `B_Number = 0;``    ``int` `cnt = 0;` `    ``while` `(N != 0) {` `        ``int` `rem = N % 2;``        ``int` `c = ``pow``(10, cnt);``        ``B_Number += rem * c;``        ``N /= 2;` `        ``// Count used to store``        ``// exponent value``        ``cnt++;``    ``}` `    ``string bin = to_string(B_Number);``    ``return` `isSuffix(``"001"``, bin);``}` `// Driver code``int` `main()``{` `    ``int` `N = 9;``    ``if` `(CheckBinaryEquivalent(N))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{``    ` `// Function returns true if``// s1 is suffix of s2``static` `boolean` `isSuffix(String s1, String s2)``{``    ``int` `n1 = s1.length();``    ``int` `n2 = s2.length();``    ` `    ``if` `(n1 > n2)``        ``return` `false``;``            ` `    ``for``(``int` `i = ``0``; i < n1; i++)``       ``if` `(s1.charAt(n1 - i - ``1``) !=``           ``s2.charAt(n2 - i - ``1``))``           ``return` `false``;``    ``return` `true``;``}``    ` `// Function to check if binary equivalent``// of a number ends in "001" or not``static` `boolean` `CheckBinaryEquivalent(``int` `N)``{``    ` `    ``// To store the binary``    ``// number``    ``int` `B_Number = ``0``;``    ``int` `cnt = ``0``;``    ` `    ``while` `(N != ``0``)``    ``{``    ` `        ``int` `rem = N % ``2``;``        ``int` `c = (``int``)Math.pow(``10``, cnt);``        ``B_Number += rem * c;``        ``N /= ``2``;``    ` `        ``// Count used to store``        ``// exponent value``        ``cnt++;``    ``}``    ``String bin = Integer.toString(B_Number);``    ``return` `isSuffix(``"001"``, bin);``}``    ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``9``;``    ` `    ``if` `(CheckBinaryEquivalent(N))``        ``System.out.println(``"Yes"``);``    ``else``        ``System.out.println(``"No"``);``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the``# above approach` `# Function returns true if``# s1 is suffix of s2``def` `isSuffix(s1, s2) :` `    ``n1 ``=` `len``(s1);``    ``n2 ``=` `len``(s2);``    ``if` `(n1 > n2) :``        ``return` `False``;``    ``for` `i ``in` `range``(n1) :``        ``if` `(s1[n1 ``-` `i ``-` `1``] !``=` `s2[n2 ``-` `i ``-` `1``]) :``            ``return` `False``;``    ``return` `True``;` `# Function to check if binary equivalent``# of a number ends in "001" or not``def` `CheckBinaryEquivalent(N) :` `    ``# To store the binary``    ``# number``    ``B_Number ``=` `0``;``    ``cnt ``=` `0``;` `    ``while` `(N !``=` `0``) :` `        ``rem ``=` `N ``%` `2``;``        ``c ``=` `10` `*``*` `cnt;``        ``B_Number ``+``=` `rem ``*` `c;``        ``N ``/``/``=` `2``;` `        ``# Count used to store``        ``# exponent value``        ``cnt ``+``=` `1``;` `    ``bin` `=` `str``(B_Number);``    ``return` `isSuffix(``"001"``, ``bin``);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``N ``=` `9``;``    ``if` `(CheckBinaryEquivalent(N)) :``        ``print``(``"Yes"``);``    ``else` `:``        ``print``(``"No"``);``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``    ` `// Function returns true if``// s1 is suffix of s2``static` `bool` `isSuffix(``string` `s1, ``string` `s2)``{``    ``int` `n1 = s1.Length;``    ``int` `n2 = s2.Length;``        ` `    ``if` `(n1 > n2)``        ``return` `false``;``                ` `    ``for``(``int` `i = 0; i < n1; i++)``       ``if` `(s1[n1 - i - 1] !=``           ``s2[n2 - i - 1])``           ``return` `false``;``    ``return` `true``;``}``        ` `// Function to check if binary equivalent``// of a number ends in "001" or not``static` `bool` `CheckBinaryEquivalent(``int` `N)``{``        ` `    ``// To store the binary``    ``// number``    ``int` `B_Number = 0;``    ``int` `cnt = 0;``        ` `    ``while` `(N != 0)``    ``{``        ``int` `rem = N % 2;``        ``int` `c = (``int``)Math.Pow(10, cnt);``        ``B_Number += rem * c;``        ``N /= 2;``        ` `        ``// Count used to store``        ``// exponent value``        ``cnt++;``    ``}``    ``string` `bin = B_Number.ToString();``    ``return` `isSuffix(``"001"``, bin);``}``    ` `// Driver code``public` `static` `void` `Main (``string``[] args)``{``    ``int` `N = 9;``    ` `    ``if` `(CheckBinaryEquivalent(N))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output

`Yes`

Time complexity: O(N)
Auxiliary space: O(1)
Efficient Approach
We can observe that the binary equivalent of a number ends in “001” only when (N – 1) is divisible by 8

Illustration:
The sequence 1, 9, 17, 25, 33……. has 001 as the suffix in their binary representation.
Nth term of the above sequence is denoted by 8 * N + 1
So the binary equivalent of a number ends in “001” only when (N – 1) % 8 == 0

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above``// approach` `#include ``using` `namespace` `std;` `// Function to check if binary``// equivalent of a number ends``// in "001" or not``bool` `CheckBinaryEquivalent(``int` `N)``{``    ``// To check if binary equivalent``    ``// of a number ends in``    ``// "001" or not``    ``return` `(N - 1) % 8 == 0;``}` `// Driver code``int` `main()``{` `    ``int` `N = 9;``    ``if` `(CheckBinaryEquivalent(N))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{``    ` `// Function to check if binary``// equivalent of a number ends``// in "001" or not``static` `boolean` `CheckBinaryEquivalent(``int` `N)``{``    ` `    ``// To check if binary equivalent``    ``// of a number ends in``    ``// "001" or not``    ``return` `(N - ``1``) % ``8` `== ``0``;``}``    ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``9``;``    ` `    ``if` `(CheckBinaryEquivalent(N))``        ``System.out.println(``"Yes"``);``    ``else``        ``System.out.println(``"No"``);``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the above approach` `# Function to check if binary``# equivalent of a number ends``# in "001" or not``def` `CheckBinaryEquivalent(N):` `    ``# To check if binary equivalent``    ``# of a number ends in``    ``# "001" or not``    ``return` `(N ``-` `1``) ``%` `8` `=``=` `0``;` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `9``;``    ` `    ``if` `(CheckBinaryEquivalent(N)):``        ``print``(``"Yes"``);``    ``else` `:``        ``print``(``"No"``);``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``        ` `// Function to check if binary``// equivalent of a number ends``// in "001" or not``static` `bool` `CheckBinaryEquivalent(``int` `N)``{``        ` `    ``// To check if binary equivalent``    ``// of a number ends in``    ``// "001" or not``    ``return` `(N - 1) % 8 == 0;``}``        ` `// Driver code``public` `static` `void` `Main (``string``[] args)``{``    ``int` `N = 9;``        ` `    ``if` `(CheckBinaryEquivalent(N))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output

`Yes`

Time complexity: O(1)
Auxiliary space: O(1)

Bitwise Approach

if any no. N has (001)2 at the end in its binary representation then N-1 has (000)2 at the end in its binary representation then doing its bitwise and with 7 (111)2 will give you (000)2 as result.

so simple return  !((N – 1) & 7)

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `// Function to check if binary equivalent of a number ends in "001" or not``bool` `CheckBinaryEquivalent(``int` `N)``{``      ``// N = 9``      ``// N - 1 = 8 => 8 = 1000``      ``// (1000 & 111) == 0 then return true``      ``// else return false``      ``return` `!((N - 1) & 7);``}` `// Driver code``int` `main()``{` `    ``int` `N = 9;``    ``if` `(CheckBinaryEquivalent(N))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG {` `  ``// Function to check if binary equivalent of a number``  ``// ends in "001" or not``  ``static` `boolean` `CheckBinaryEquivalent(``int` `N)``  ``{` `    ``// N = 9``    ``// N - 1 = 8 => 8 = 1000``    ``// (1000 & 111) == 0 then return true``    ``// else return false``    ``if` `(((N - ``1``) & ``7``) > ``0``)``      ``return` `false``;``    ``else``      ``return` `true``;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `N = ``9``;` `    ``if` `(CheckBinaryEquivalent(N))``      ``System.out.println(``"Yes"``);``    ``else``      ``System.out.println(``"No"``);``  ``}``}` `// This code is contributed by ajaymakvana`

## Python3

 `# Python implementation of the above approach` `# Function to check if binary equivalent of a number ends in "001" or not``def` `CheckBinaryEquivalent(N):``  ` `        ``# N = 9``        ``# N - 1 = 8 => 8 = 1000``        ``# (1000 & 111) == 0 then return true``        ``# else return false``    ``return` `(``not``((N ``-` `1``) & ``7``))` `N ``=` `9``if` `(CheckBinaryEquivalent(N)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `    ``# This code is contributed by ajaymakvava.`

## C#

 `// C# implementation of the above approach``using` `System;``class` `GFG {` `  ``// Function to check if binary equivalent of a number ends in "001" or not``  ``static` `bool` `CheckBinaryEquivalent(``int` `N)``  ``{` `    ``// N = 9``    ``// N - 1 = 8 => 8 = 1000``    ``// (1000 & 111) == 0 then return true``    ``// else return false``    ``if` `(((N - 1) & 7) > 0)``      ``return` `false``;``    ``else``      ``return` `true``;``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int` `N = 9;` `    ``if` `(CheckBinaryEquivalent(N))``      ``Console.WriteLine(``"Yes"``);``    ``else``      ``Console.WriteLine(``"No"``);``  ``}``}` `// This code is contributed by ajaymakvana`

## Javascript

 `// JavaScript implementation of the above approach` `// Function to check if binary equivalent of a number ends in "001" or not``function` `CheckBinaryEquivalent(N)``{``      ``// N = 9``      ``// N - 1 = 8 => 8 = 1000``      ``// (1000 & 111) == 0 then return true``      ``// else return false``      ``return` `!((N - 1) & 7);``}` `// Driver code``let N = 9;``if` `(CheckBinaryEquivalent(N))``    ``console.log(``"Yes"``);``else``    ``console.log(``"No"``);` `// This code is contributed by phasing17`

Output

`Yes`
```Time Complexity: O(1)
Space COmplexity: O(1)```

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