Dado un entero positivo N , la tarea es verificar si el equivalente binario de ese entero termina en «001» o no.
Escriba “ Sí ” si termina en “001”. De lo contrario, escriba “ No ”.
Ejemplos:
Entrada : N = 9
Salida : Sí
Explicación
Binario de 9 = 1001, que termina en 001
Entrada : N = 5
Salida : No
Binario de 5 = 101, que no termina en 001
Enfoque ingenuo
Encuentre el equivalente binario de N y verifique si 001 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 "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>
Producción
Yes
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Enfoque eficiente
Podemos observar que el equivalente binario de un número termina en “001” solo cuando (N – 1) es divisible por 8 .
Ilustración:
La secuencia 1, 9, 17, 25, 33……. tiene 001 como sufijo en su representación binaria. El término
N de la secuencia anterior se denota por 8 * N + 1. Por lo tanto, el equivalente binario de un número termina en «001» solo cuando (N – 1) % 8 == 0
A continuación se muestra la implementación del enfoque anterior:
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>
Producción
Yes
Complejidad temporal: O(1)
Espacio auxiliar: O(1)
Enfoque bit a bit
si alguno no N tiene (001) 2 al final en su representación binaria, luego N-1 tiene (000) 2 al final en su representación binaria, luego haciendo bit a bit y con 7 (111) 2 le dará (000) 2 como resultado.
tan simple retorno !((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
Producción
Yes
Time Complexity: O(1) Space COmplexity: O(1)