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Check whether the binary equivalent of a number ends with given string or not

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  • Last Updated : 08 Jun, 2022
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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


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