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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the
// above approach
  
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above
// approach
  
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

Yes

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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01