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

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

Last Updated : 18 Jun, 2020

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 <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; 
} 

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 

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.
N^th 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 <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; 
} 

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

Output:

Yes

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

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My Personal Notes

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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