Given a number which may be in octal or in decimal. If the number is not octal then convert it into octal then check if it is palindrome or not. If it is palindrome then print 1 otherwise print 0. If the number is already in octal then check if the number is palindrome or not.
Examples :
Input :n = 111 Output : Yes Explanation: all digits of 111 are in 0-7 so it is a octal number. Read 111 the result is same backward and forward so it is a palindrome number. Input : 68 Output : No Explanation: 68 is not a octal number because it's all digits are not in 0-7. So first we need to convert it into octal (68)base10(Decimal) = (104)base8(octal) 104 is not palindrome. Input : 97 Output : Yes Explanation: 97 is not a octal number because it's all digits are not in 0-7 so first we need to convert it into decimal to octal (97)base10(Decimal) = (141)base8(octal) 141 is palindrome so output = 1.
Octal Number : The octal numeral system, or oct for short, is the base-8 number system, and uses the digits 0 to 7.
How to Convert Decimal Number to Octal Number :
C++
// C++ program to check if octal// representation of a number is prime#include <iostream>using namespace std;const int MAX_DIGITS = 20;/* Function to Check no is in octalor not */bool isOctal(long n){ while (n) { if ((n % 10) >= 8) return false; else n = n / 10; } return true;}/* Function To check no is palindromeor not*/int isPalindrome(long n){ // If number is already in octal, we traverse // digits using repeated division with 10. Else // we traverse digits using repeated division // with 8 int divide = (isOctal(n) == false)? 8 : 10; // To store individual digits int octal[MAX_DIGITS]; // Traversing all digits int i = 0; while (n != 0) { octal[i++] = n % divide; n = n / divide; } // checking if octal no is palindrome for (int j = i - 1, k = 0; k <= j; j--, k++) if (octal[j] != octal[k]) return false; return true;}// Driver codeint main(){ long n = 97; if (isPalindrome(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to check if // octal representation of // a number is primeimport java.io.*;import java.util.*;import java.lang.*;class GFG{static int MAX_DIGITS = 20;/* Function to Check no is in octal or not */static int isOctal(int n){ while (n > 0) { if ((n % 10) >= 8) return 0; else n = n / 10; } return 1;}/* Function To check no is palindrome or not*/static int isPalindrome(int n){ // If number is already in // octal, we traverse digits // using repeated division // with 10. Else we traverse // digits using repeated // division with 8 int divide = (isOctal(n) == 0) ? 8 : 10; // To store individual digits int octal[] = new int[MAX_DIGITS]; // Traversing all digits int i = 0; while (n != 0) { octal[i++] = n % divide; n = n / divide; } // checking if octal // no is palindrome for (int j = i - 1, k = 0; k <= j; j--, k++) if (octal[j] != octal[k]) return 0; return 1;}// Driver Codepublic static void main(String[] args){ int n = 97; if (isPalindrome(n) > 0) System.out.print("Yes"); else System.out.print("No");}}// This code is contributed // by Akanksha Rai(Abby_akku) |
Python3
# Python3 program to check if # octal representation of # a number is primeMAX_DIGITS = 20;# Function to Check no # is in octal or notdef isOctal(n): while(n): if((n % 10) >= 8): return False else: n = int(n / 10) return True# Function To check no # is palindrome or notdef isPalindrome(n): # If number is already in # octal, we traverse digits # using repeated division # with 10. Else we traverse # digits using repeated # division with 8 divide = 8 if(isOctal(n) == False) else 10 # To store individual digits octal=[] # Traversing all digits while (n != 0): octal.append(n % divide) n = int(n / divide) # checking if octal # no is palindrome j = len(octal)-1 k = 0 while(k <= j): if(octal[j] != octal[k]): return False j-=1 k+=1 return True# Driver Codeif __name__=='__main__': n = 97; if (isPalindrome(n)): print("Yes") else: print("No")# This code is contributed by mits |
C#
// C# program to check if // octal representation of // a number is primeusing System;class GFG{static long MAX_DIGITS = 20;/* Function to Check no is in octal or not */static long isOctal(long n){ while (n > 0) { if ((n % 10) >= 8) return 0; else n = n / 10; } return 1;}/* Function To check no is palindrome or not*/static long isPalindrome(long n){ // If number is already in // octal, we traverse digits // using repeated division // with 10. Else we traverse // digits using repeated // division with 8 long divide = (isOctal(n) == 0) ? 8 : 10; // To store individual digits long[] octal = new long[MAX_DIGITS]; // Traversing all digits long i = 0; while (n != 0) { octal[i++] = n % divide; n = n / divide; } // checking if octal // no is palindrome for (long j = i - 1, k = 0; k <= j; j--, k++) if (octal[j] != octal[k]) return 0; return 1;}// Driver Codestatic int Main(){ long n = 97; if (isPalindrome(n) > 0) Console.Write("Yes"); else Console.Write("No"); return 0;}}// This code is contributed // by mits |
PHP
<?php// PHP program to check if // octal representation of // a number is prime$MAX_DIGITS = 20;// Function to Check no // is in octal or notfunction isOctal($n){ while ($n) { if (($n % 10) >= 8) return false; else $n = (int)$n / 10; } return true;}// Function To check no // is palindrome or notfunction isPalindrome($n){ global $MAX_DIGITS; // If number is already in // octal, we traverse digits // using repeated division // with 10. Else we traverse // digits using repeated // division with 8 $divide = (isOctal($n) == false) ? 8 : 10; // To store individual digits $octal; // Traversing all digits $i = 0; while ($n != 0) { $octal[$i++] = $n % $divide; $n = (int)$n / $divide; } // checking if octal // no is palindrome for ($j = $i - 1, $k = 0; $k <= $j; $j--, $k++) if ($octal[$j] != $octal[$k]) return -1; return 0;}// Driver Code$n = 97;if (isPalindrome($n))echo "Yes";elseecho "No";// This code is contributed by ajit?> |
Javascript
<script>// Javascript program to check if octal // representation of a number is prime const MAX_DIGITS = 20; /* Function to Check no is in octal or not */function isOctal(n) { while (n) { if ((n % 10) >= 8) return false; else n = Math.floor(n / 10); } return true; } /* Function To check no is palindrome or not*/function isPalindrome(n) { // If number is already in octal, we traverse // digits using repeated division with 10. Else // we traverse digits using repeated division // with 8 var divide = (isOctal(n) == false)? 8 : 10; // To store individual digits var octal = new Array(MAX_DIGITS); // Traversing all digits var i = 0; while (n != 0) { octal[i++] = n % divide; n = Math.floor(n / divide); } // checking if octal no is palindrome for (var j = i - 1, k = 0; k <= j; j--, k++) if (octal[j] != octal[k]) return false; return true; } // Driver code var n = 97; if (isPalindrome(n)) document.write("Yes"); else document.write("No"); // This code is contributed by Mayank Tyagi</script> |
Output :
Yes
Time complexity: O(logn) where n is given input number
Auxiliary Space: O(1)
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