Python Program to Check if a String is Palindrome or Not

Last Updated : 18 Aug, 2023

Given a string, write a python function to check if it is palindrome or not. A string is said to be a palindrome if the reverse of the string is the same as the string. For example, “radar” is a palindrome, but “radix” is not a palindrome.

Examples:

Input : malayalam
Output : Yes

Input : geeks
Output : No

Python Program to Check if a String is Palindrome or Not Using Native Approach

Here we will find reverse of the string and then Check if reverse and original are same or not.

Python

# function which return reverse of a string
 
def isPalindrome(s):
    return s == s[::-1]
 
 
# Driver code
s = "malayalam"
ans = isPalindrome(s)
 
if ans:
    print("Yes")
else:
    print("No")

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(1)

Check if a String is Palindrome or Not Using Iterative Method

Run a loop from starting to length/2 and check the first character to the last character of the string and second to second last one and so on …. If any character mismatches, the string wouldn’t be a palindrome.

Below is the implementation of above approach:

Python

# function to check string is
# palindrome or not
def isPalindrome(str):
 
    # Run loop from 0 to len/2
    for i in range(0, int(len(str)/2)):
        if str[i] != str[len(str)-i-1]:
            return False
    return True
 
# main function
s = "malayalam"
ans = isPalindrome(s)
 
if (ans):
    print("Yes")
else:
    print("No")

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(1)

Check String is Palindrome or Not Using the inbuilt function to reverse a string

In this method, the predefined function ‘ ‘.join(reversed(string)) is used to reverse string.

Below is the implementation of the above approach:

Python

# function to check string is
# palindrome or not
def isPalindrome(s):
 
    # Using predefined function to
    # reverse to string print(s)
    rev = ''.join(reversed(s))
 
    # Checking if both string are
    # equal or not
    if (s == rev):
        return True
    return False
 
# main function
s = "malayalam"
ans = isPalindrome(s)
 
if (ans):
    print("Yes")
else:
    print("No")

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(n)

Check String is Palindrome using one extra variable

In this method, the user takes a character of string one by one and store it in an empty variable. After storing all the characters user will compare both the string and check whether it is palindrome or not.

Python

# Python program to check
# if a string is palindrome
# or not
 
x = "malayalam"
 
w = ""
for i in x:
    w = i + w
 
if (x == w):
    print("Yes")
else:
    print("No")

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(n)

Check String is Palindrome using flag

In this method, the user compares each character from starting and ending in a for loop and if the character does not match then it will change the status of the flag. Then it will check the status of the flag and accordingly and print whether it is a palindrome or not.

Python

# Python program to check
# if a string is palindrome
# or not
st = 'malayalam'
j = -1
flag = 0
for i in st:
    if i != st[j]:
        flag = 1
        break
    j = j - 1
if flag == 1:
    print("NO")
else:
    print("Yes")

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(1)

Check String is Palindrome using recursion

This method compares the first and the last element of the string and gives the rest of the substring to a recursive call to itself.

Python3

# Recursive function to check if a
# string is palindrome
def isPalindrome(s):
 
    # to change it the string is similar case
    s = s.lower()
    # length of s
    l = len(s)
 
    # if length is less than 2
    if l < 2:
        return True
 
    # If s[0] and s[l-1] are equal
    elif s[0] == s[l - 1]:
 
        # Call is palindrome form substring(1,l-1)
        return isPalindrome(s[1: l - 1])
 
    else:
        return False
 
# Driver Code
s = "MalaYaLam"
ans = isPalindrome(s)
 
if ans:
    print("Yes")
 
else:
    print("No")

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(n)

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