Dado un entero positivo N , la tarea es verificar si el equivalente binario de ese entero termina con la string dada str o no.
Escriba «Sí» si termina en «str». De lo contrario, escriba “No”.
Ejemplos :
Entrada: N = 23, str = “111”
Salida: Sí
Explicación:
Binario de 23 = 10111, que termina en “111”Entrada: N = 5, str = “111”
Salida: No
Enfoque : la idea es encontrar el equivalente binario de N y verificar si str es un sufijo de su equivalente binario.
A continuación se muestra la implementación del enfoque anterior:
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>
Producción:
Yes
Complejidad de tiempo: O(|str|)
Espacio auxiliar: O (log 2 (N)) ya que estamos creando un contenedor de strings adicional para convertir N en su string binaria equivalente