Answer:
Program is as follow:
Explanation:
#include<iostream.h>
#include < conio.h>
bool isVowel(char b)
{
return
( b=='a' || b=='A' || b=='e' || b=='E' || b=='i' || b=='I' || b=='o' || b=='O' || b== 'u' || b=='U');
For swaoing the vowels
string reverseVowel ( String str)
{
int j = 0;
string v ;
for ( int i= 0 ; str [i]]!='0'; i++)
if( isv(str[i]))
v[j++] = str[i[;
for ( int i = 0; str[i]! = '0' ; i++
if ( isv (str[i]))
str{i} = v [--j]
return str;
}
int main ()
{ string str = " Programming";
cout<<swapv(str);
return 0 ;
}
Answer:
(s.charAt(0) != s.charAt(s.length()-1))
Explanation:
public class Palindrome
{
public static void main(String[] args) {
System.out.println(isPalindrome("car"));
System.out.println(isPalindrome("carac"));
}
public static boolean isPalindrome(String s) {
if (s.length() <= 1)
return true;
else if (s.charAt(0) != s.charAt(s.length()-1))
return false;
else
return isPalindrome(s.substring(1, s.length() - 1));
}
}
You may see the whole code above with two test scenarios.
The part that needs to be filled is the base case that if the character at position 0 is not equal to the character at the last position in the string, that means the string is not a palindrome. (If this condition meets, it checks for the second and the one before the last position, and keeps checking)
Answer:
ummmm try the inequality protragathron theorum
Explanation:
ok
def recursiveFactorial(number):
if number > 1:
return number * recursiveFactorial(number-1)
else:
return 1
stringNum = input("Enter a positive integer: ")
num = int(stringNum)
print(recursiveFactorial(num))
