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

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  • Last Updated : 19 Jul, 2022
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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 <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

Javascript




<script>
 
// javascript implementation of the above approach
 
     
// Function returns true if
// s1 is suffix of s2
function isSuffix( s1,  s2)
{
    var n1 = s1.length;
    var n2 = s2.length;
         
    if (n1 > n2)
        return false;
                 
    for(var 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
function CheckBinaryEquivalent( N)
{
         
    // To store the binary
    // number
    var B_Number = 0;
    var cnt = 0;
         
    while (N != 0)
    {
        var rem = N % 2;
        var c = Math.pow(10, cnt);
        B_Number += rem * c;
        N = Math.floor(N/ 2);
         
        // Count used to store
        // exponent value
        cnt++;
    }
     console.log(B_Number);
    var bin = B_Number.toString();
    return isSuffix("001", bin);
}
     
// Driver code
 
    var N = 9;
     
    if (CheckBinaryEquivalent(N))
        document.write("Yes");
    else
        document.write("No");
         
         
</script>

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 <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

Javascript




<script>
 
// Javascript implementation of the above
// approach
 
// Function to check if binary
// equivalent of a number ends
// in "001" or not
function CheckBinaryEquivalent(N)
{
    // To check if binary equivalent
    // of a number ends in
    // "001" or not
    return (N - 1) % 8 == 0;
}
 
// Driver code
var N = 9;
if (CheckBinaryEquivalent(N))
    document.write( "Yes");
else
    document.write( "No");
 
</script>

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 <bits/stdc++.h>
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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