Check whether the binary equivalent of a number ends with given string or not
Given a positive integer N, the task is to check whether the binary equivalent of that integer ends with the given string str or not.
Print “Yes” if it ends in “str”. Otherwise, Print “No”.
Examples:
Input: N = 23, str = “111”
Output: Yes
Explanation:
Binary of 23 = 10111, which ends with “111”Input: N = 5, str = “111”
Output: No
Approach: The idea is to find the Binary Equivalent of N and check if str 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 "111" or not bool CheckBinaryEquivalent( int N, string str) { // 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(str, bin); } // Driver code int main() { int N = 23; string str = "111" ; if (CheckBinaryEquivalent(N, str)) 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(), 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 "111" or not static boolean CheckBinaryEquivalent( int N, String str) { // 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(str, bin); } // Driver code public static void main(String[] args) { int N = 23 ; String str = "111" ; if (CheckBinaryEquivalent(N, str)) System.out.print( "Yes\n" ); else System.out.print( "No\n" ); } } // This code is contributed by shubham |
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 "111" or not def CheckBinaryEquivalent(N, s): # To store the binary # number B_Number = 0 ; cnt = 0 ; while (N ! = 0 ): rem = N % 2 ; c = pow ( 10 , cnt); B_Number + = rem * c; N / / = 2 ; # Count used to store # exponent value cnt + = 1 ; bin = str (B_Number); return isSuffix(s, bin ); # Driver code if __name__ = = "__main__" : N = 23 ; s = "111" ; if (CheckBinaryEquivalent(N, s)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by rutvik_56 |
C#
// C# implementation of the // above approach using System; using System.Collections; class GFG{ // Function returns true if // s1 is suffix of s2 static bool isSuffix(String s1, String s2) { int n1 = s1.Length, 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 "111" or not static bool CheckBinaryEquivalent( int N, String str) { // 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(str, bin); } // Driver Code public static void Main (String[] args) { int N = 23; String str = "111" ; if (CheckBinaryEquivalent(N, str)) Console.WriteLine( "Yes\n" ); else Console.WriteLine( "No\n" ); } } // This code is contributed by jana_sayantan |
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 "111" or not function CheckBinaryEquivalent(N, str) { // 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 = parseInt(N/2); // Count used to store // exponent value cnt++; } var bin = B_Number.toString(); return isSuffix(str, bin); } // Driver code var N = 23; var str = "111" ; if (CheckBinaryEquivalent(N, str)) document.write( "Yes" ); else document.write( "No" ); </script> |
Output:
Yes
Time Complexity: O(|str|)
Auxiliary Space: O(log2(N)) as we are making an extra string bin for converting N into its binary equivalent string