Dados dos enteros A y B , la tarea es verificar si el entero A es una rotación de los dígitos del entero B o no. Si se encuentra que es cierto, escriba «Sí» . De lo contrario, escriba “No” .
Ejemplos:
Entrada: A= 976, B= 679
Salida: SíEntrada: A= 974, B= 2345
Salida: No
Enfoque: siga los pasos a continuación para resolver el problema:
- Si A == B , imprima «Sí» .
- Calcule el recuento de dígitos del entero presente en el entero A y B en las variables, digamos dig1 y dig2 .
- Si se encuentra que dig1 != dig2 no es verdadero, entonces imprima “No” .
- Inicialice una variable, digamos temp . Asigne temp = A .
- Ahora, itera y realiza las siguientes operaciones:
- Imprime “Sí” si A es igual a B y rompe.
- Compruebe si (A== temp), es decir , A se vuelve igual a la temperatura entera original . Si se encuentra que es cierto, escriba «No» y rompa.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the integer
// A is a rotation of the integer B
int check(int A, int B)
{
if (A == B) {
return 1;
}
// Stores the count of digits in A
int dig1 = floor(log10(A) + 1);
// Stores the count of digits in B
int dig2 = floor(log10(B) + 1);
// If dig1 not equal to dig2
if (dig1 != dig2) {
return 0;
}
int temp = A;
while (1) {
// Stores position of first digit
int power = pow(10, dig1 - 1);
// Stores the first digit
int firstdigit = A / power;
// Rotate the digits of the integer
A = A - firstdigit * power;
A = A * 10 + firstdigit;
// If A is equal to B
if (A == B) {
return 1;
}
// If A is equal to the initial
// value of integer A
if (A == temp) {
return 0;
}
}
}
// Driver Code
int main()
{
int A = 967, B = 679;
if (check(A, B))
cout << "Yes";
else
cout << "No" << endl;
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG {
// Function to check if the integer
// A is a rotation of the integer B
static int check(int A, int B)
{
if (A == B) {
return 1;
}
// Stores the count of digits in A
int dig1 = (int)Math.floor(Math.log10(A) + 1);
// Stores the count of digits in B
int dig2 = (int)Math.floor(Math.log10(B) + 1);
// If dig1 not equal to dig2
if (dig1 != dig2) {
return 0;
}
int temp = A;
while (true) {
// Stores position of first digit
int power = (int)Math.pow(10, dig1 - 1);
// Stores the first digit
int firstdigit = A / power;
// Rotate the digits of the integer
A = A - firstdigit * power;
A = A * 10 + firstdigit;
// If A is equal to B
if (A == B) {
return 1;
}
// If A is equal to the initial
// value of integer A
if (A == temp) {
return 0;
}
}
}
// Driver Code
public static void main(String[] args)
{
int A = 967, B = 679;
if (check(A, B) == 1)
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Dharanendra L V.
Python3
# Python3 implementation of the approach
import math
# Function to check if the integer
# A is a rotation of the integer B
def check(A, B) :
if (A == B) :
return 1
# Stores the count of digits in A
dig1 = math.floor(math.log10(A) + 1)
# Stores the count of digits in B
dig2 = math.floor(math.log10(B) + 1)
# If dig1 not equal to dig2
if (dig1 != dig2) :
return 0
temp = A
while (True) :
# Stores position of first digit
power = pow(10, dig1 - 1)
# Stores the first digit
firstdigit = A // power
# Rotate the digits of the integer
A = A - firstdigit * power
A = A * 10 + firstdigit
# If A is equal to B
if (A == B) :
return 1
# If A is equal to the initial value of integer A
if (A == temp) :
return 0
# Driver code
A, B = 967, 679
if (check(A, B)) :
print("Yes")
else :
print("No")
# This code is contributed by divyesh072019.
C#
// C# implementation of the approach
using System;
public class GFG
{
// Function to check if the integer
// A is a rotation of the integer B
static int check(int A, int B)
{
if (A == B)
{
return 1;
}
// Stores the count of digits in A
int dig1 = (int)Math.Floor(Math.Log10(A) + 1);
// Stores the count of digits in B
int dig2 = (int)Math.Floor(Math.Log10(B) + 1);
// If dig1 not equal to dig2
if (dig1 != dig2)
{
return 0;
}
int temp = A;
while (true)
{
// Stores position of first digit
int power = (int)Math.Pow(10, dig1 - 1);
// Stores the first digit
int firstdigit = A / power;
// Rotate the digits of the integer
A = A - firstdigit * power;
A = A * 10 + firstdigit;
// If A is equal to B
if (A == B)
{
return 1;
}
// If A is equal to the initial
// value of integer A
if (A == temp)
{
return 0;
}
}
}
// Driver Code
public static void Main(String[] args)
{
int A = 967, B = 679;
if (check(A, B) == 1)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation of the approach
// Function to check if the integer
// A is a rotation of the integer B
function check(A, B)
{
if (A == B)
{
return 1;
}
// Stores the count of digits in A
let dig1 = Math.floor(Math.log10(A) + 1);
// Stores the count of digits in B
let dig2 = Math.floor(Math.log10(B) + 1);
// If dig1 not equal to dig2
if (dig1 != dig2)
{
return 0;
}
let temp = A;
while (true)
{
// Stores position of first digit
let power = Math.pow(10, dig1 - 1);
// Stores the first digit
let firstdigit = parseInt(A / power, 10);
// Rotate the digits of the integer
A = A - firstdigit * power;
A = A * 10 + firstdigit;
// If A is equal to B
if (A == B)
{
return 1;
}
// If A is equal to the initial
// value of integer A
if (A == temp)
{
return 0;
}
}
}
let A = 967, B = 679;
if (check(A, B) == 1)
document.write("Yes");
else
document.write("No");
// This code is contributed by suresh07.
</script>
Producción:
Yes
Complejidad de tiempo: O(dígito(N))
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por patelajeet y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA