6

I have a code to check whether a word is palindrome or not:

str = input("Enter the string") l = len(str) p = l-1 index = 0 while index < p: if str[index] == str[p]: index = index + 1 p = p-1 print("String is a palindrome") break else: print("string is not a palindrome")

If a word is inputted, for example : rotor , I want the program to check whether this word is palindrome and give output as "The given word is a palindrome".

But I'm facing problem that, the program checks first r and r and prints "The given word is a palindrome" and then checks o and o and prints "The given word is a palindrome". It prints the result as many times as it is checking the word.

I want the result to be delivered only once. How to change the code?

Improve this question
2
  • Is this an exercise where you have to do it using a while, or can you use the simpler, if word == word[::-1] ? (where word is your input) Commented Aug 22, 2017 at 18:25
  • 1
    Possible duplicate of How to check for palindrome using Python logic Commented Aug 22, 2017 at 18:29

9 Answers 9

Reset to default
8

Just reverse the string and compare it against the original

string_to_check = input("Enter a string") if string_to_check == string_to_check[::-1]: print("This is a palindrome") else: print("This is not a palindrome")
Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

2

I had to make a few changes in your code to replicate the output you said you saw.

Ultimately, what you want is that the message be displayed only at the end of all the comparisons. In your case, you had it inside the loop, so each time the loop ran and it hit the if condition, the status message was printed. Instead, I have changed it so that it only prints when the two pointers index and p are at the middle of the word.

str = input("Enter the string: ") l = len(str) p = l-1 index = 0 while index < p: if str[index] == str[p]: index = index + 1 p = p-1 if index == p or index + 1 == p: print("String is a palindrome") else: print("string is not a palindrome") break
Improve this answer

5 Comments

Thanks bro. But I have one doubt, why did used - "index + 1 == p" but not "index == p" in if index == p or index + 1 == p:
There can be 2 types of palindromes. Something like "rotor", which has an odd number of letters, and so index and p would converge on the middle letter. But with other words like "maam", index and p would end up on the two 'a's in the middle. Hence, to find that end point, we use index + 1 == p.
This is why you should use raw_input instead of input.
But raw_input is removed and it is same as input in python 3 right?
Oh haha, sorry, didn't notice that you used Python 3. I'll edit the code.
1

I see that most of the solutions in the internet is where either the string is taken as an input or the string is just reversed and then tested. Below is a solution that considers two points: 1) The input is a very large string therefore, it cannot be just taken from a user. 2) The input string will have capital casing and special characters. 3) I have not looked into the complexity and see further improvement. Would invite suggestions.

 def isPalimdromeStr(self, strInput): strLen = len(strInput) endCounter = strLen - 1 startCounter = 0 while True: # print(startCounter, endCounter) # print(strInput[startCounter].lower(), strInput[endCounter].lower()) while not blnValidCh(self, strInput[startCounter].lower()): startCounter = startCounter + 1 while not blnValidCh(self, strInput[endCounter].lower()): endCounter = endCounter - 1 # print(startCounter, endCounter) # print(strInput[startCounter].lower() , strInput[endCounter].lower()) if (strInput[startCounter].lower() != strInput[endCounter].lower()): return False else: startCounter = startCounter + 1 endCounter = endCounter - 1 if (startCounter == strLen - 1): return True # print("---") def blnValidCh(self, ch): if rePattern.match(ch): return True return False global rePattern rePattern = re.compile('[a-z]') blnValidPalindrome = classInstance.isPalimdromeStr("Ma##!laYal!! #am") print("***") print(blnValidPalindrome)
Improve this answer

Comments

0 
[::-1]

This code is the easiest way of reversing a string. A palindrome is something that reads the same regardless of what side you're reading it from. So the easiest function to check whether or not the word/string is recursive or not would be:

def checkPalindrome(s): r = s[::-1] if s == r: return True else: return False

Now you can check this code. Eg:

checkPalindrome("madam") >>> True checkPalindrome("sentence") >>> False
Improve this answer

Comments

0

Change your implementation to the following one:

string_to_be_checked = input("enter your string ") if string_to_be_checked == string_to_be_checked[::-1] : print("palindrome") else: print("not palindrome")
Improve this answer

Comments

0

the easiest way is

str = input('Enter the string: ') if str == str[::-1]: print('Palindrome') else print('Not')

the same login can also be applied to number palindrome

Improve this answer

Comments

0 
def palindrome(s): if len(s)<1: return True else: if s[0]==s[-1]: return palindrome(s[1:-1]) else: return False a=str(input("enter data") if palindrome(a) is True: print("string is palindrome") else: print("string is not palindrome")
Improve this answer

Comments

0 
string = 'ASDBDSA' if string[:]==string[::-1]: print("palindrome") else: print("not palindrome")
Improve this answer

Comments

0

You can also do it by using ternary operators in python.

string = "Madam" print("palindrome") if string == string[::-1] else print("not a palindrome")
Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.