Check if a number is palindrome
Given an alphabetical string ‘S’. Determine whether it is palindrome or not. A palindrome is a string that is equal to itself upon reversing it.

For example:

‘S’ = racecar The reverse of ‘S’ is: racecar Since ‘S’ is equal to its reverse. So ‘S’ is a palindrome. Hence output will be 1.

Input Format:

The first line of the input contains a single integer ‘T’ representing the no. of test cases. The first line of each test case contains a single alphabetical string, ‘S’.

Output Format:

For each test case, print a single integer value 1 if the given string ‘S’ is palindrome and 0 otherwise. Print a separate line for each test case.

Note

You are not required to print anything; it has already been taken care of. Just implement the function and return the answer.

Constraints -

1 ≤ T ≤ 1000 1 ≤ |S| ≤ 10^5 S consists of only lowercase english alphabets. Σ|S| ≤ 2 * 10^6 Time limit: 1 Sec

Question

Check if a number is palindrome

Given an alphabetical string ‘S’. Determine whether it is palindrome or not. A palindrome is a string that is equal to itself upon reversing it.

For example:

‘S’ = racecar
The reverse of ‘S’ is: racecar
Since ‘S’ is equal to its reverse. So ‘S’ is a palindrome.
Hence output will be 1.

Input Format:

The first line of the input contains a single integer ‘T’ representing the no. of test cases.

The first line of each test case contains a single alphabetical string, ‘S’.

Output Format:

For each test case, print a single integer value 1 if the given string ‘S’ is palindrome and 0 otherwise.

Print a separate line for each test case.

Note

You are not required to print anything; it has already been taken care of. Just implement the function and return the answer.

Constraints -

1 ≤ T ≤ 1000
1 ≤ |S| ≤ 10^5
S consists of only lowercase english alphabets.
Σ|S| ≤ 2 * 10^6

Time limit: 1 Sec

Accepted Answer

One approach could be to first reverse digits of n, then compare the reverse of n with n. If both are same, then return true, else false.

Pseudo code :

reverseDigits(num)
{
	Initialise a variable rev_num to 0
	while (num is greater than 0) {
		rev_num = rev_num * 10 + num % 10
		num = num / 10
	}
		return rev_num
}

/* Function to check if n is Palindrome*/
isPalindrome(n)
{
	// get the reverse of n
	rev_n = reverseDigits(n)

	// Check if rev_n and n are same or not.
	if (rev_n == n)
		return 1
	else
		return 0
}

Accepted Answer

1Space Complexity: O(1)Explanation: Time Complexity: O(n)Explanation:
Java (SE 1.8)

/*
  Time Complexity: O( N ).
  Space Complexity: O( 1 ).

  Where N is the length of the string.
*/

public class Solution {

    public static Boolean isPalindrome(String s) {
        int n = s.length();
        int i;

        /*
         * For each character in the first half check its
         * corresponding element in the second half of the string.
         */
        for (i = 0; i < n / 2; i++) {
            if (s.charAt(i) != s.charAt(n - 1 - i)) {
                break;
            }
        }

        // If we broke early from the loop then it isn't a palindrome.
        return (i == n / 2);
    }
}

C++ (g++ 5.4)

/*
    Time Complexity: O(N) 
    Space Complexity: O(1) 

    Where 'N' is the length of the given string.
*/

// Given Function
bool isPalindrome(string &s)
{
	int n = s.size(), i;

	/* For each character in the first half check its 
	   corresponding element in the second half of the string.
    */
	for(i = 0; i < n/2; i++) {
		if(s[i] != s[n-1-i]) {
			break;
		}
	}

	// If we broke early from the loop then it isn't a palindrome.
	return (i == n/2);
}

Python (3.5)

'''
    Time Complexity: O(N) 
    Space Complexity: O(1) 

    Where 'N' is the length of the given string.
'''

# Given Function
def isPalindrome(s: str) -> bool:

    n = len(s)

    '''
        For each character in the first half check its 
        corresponding element in the second half of the string.
    '''
    for i in range(n // 2):
        if s[i] != s[n-1-i]:
            break

	# If we broke early from the loop then it isn't a palindrome.
    return (i == n // 2 - 1)

Given an alphabetical string ‘S’. Determine whether it is palindrome or not. A palindrome is a string that is equal to itself upon reversing it.

For example:

Input Format:

Output Format:

Note

Constraints -

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