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 001Input: 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 |
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 |
Yes
Time complexity: O(1)
Auxiliary space: O(1)
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