Get nth Prime and Palindrome number-CodePudding

def ispalindrome(number):
    x=number
    reverse=0
    while(x!=0):
        digit=x
        reverse=reverse*10 digit
        x=x//10
    if number==reverse:
        return True
    else:
        return False
def isprime(num):
    y=True
    for i in range(2,num):
        if(num%i==0):
            y= False
            break
    return y
n=int(input("Enter the value of n:"))
primeAndPalindrome=[]
p=10**n
for j in range (2,p):
    if ispalindrome(j) and isprime(j):
        primeAndPalindrome.append(j)
    if len(primeAndPalindrome)==n 10:
        break
print(f'{n}th Prime Palindrome number is {primeAndPalindrome[n-1]}')

if len(primeAndPalindrome)==n 10:

what is use of this line , if I don't use this my code runs but the execution takes much time.

Can someone explain the use of this line.

CodePudding user response：

You are checking the nth prime and palindrome in a range of numbers from 2 to 10**n. The bigger n input is the bigger your list and the more time it will take to execute. if len(primeAndPalindrome)==n 10: limits your list size to n 10 only so it seems faster but the problem is still your program cannot handle big numbers.

Also, your checking functions (isPalindrome and isPrime) could be improved a lot.

CodePudding user response：

As @Florian EDEMESSI says the approach needs improving so here are re-coded check functions to start with:

def ispalindrome(number):
    x=str(number)
    return x == x[::-1]


def isprime(num):
    if num <= 1:
        return False
    for i in range(2,int(num**0.5)):
        if(num%i==0):
            return False
    return True