Approach

Written by Winnie Zhang from CSESoc Education

This question is all about problem solving and understanding strings/arrays!

For this question, we need to firstly read in the input. To do this, we can use fgets to help read in the line of input. Luckily, the input does not have a newline character at the end so we do not need to override it with a \0. We can see the syntax of using fgets in this question below:

#define MAX 4096
...
// Variable `str` will store the word
char str[MAX];
// Take in the word from standard input (the terminal) and put it in the `str` variable
fgets(str, MAX, stdin);
int len = strlen(str);

As a side note, fgets reads a string until it encounters \n (or the MAX characters is reached or it reads EOF) so if you ran the code locally, you might need to consider the stripping away \n if you press the enter key after typing the input. To do this, you would find the index in which the newline was located (which would be len - 1 because the strlen function counts \n as a character in the string's length), then override it with a null terminator (would look something like str[len - 1] = '\0').

Now comes the harder logic. How do we know it is a palindrome?

Well we want to ensure that the first letter is the same as the last letter, the second letter is the same as the second last letter and so on. This sounds like repetitive work, like a loop!

Let us take the first case, check if the first letter is the same as the last letter. Here, we would want to see if index 0 of the array is the same as the last index of the array. The last index of the array isn't len of the array because indexing starts at 0, not 1, so the last index is len - 1.

So if we check that they weren't the same letter, we can immediately say that this is not a palindrome and early return:

if (str[0] != str[len - 1]) {
    // print that it is not a palindrome
    return 0;
}

Alternatively, you can have what we call a "flag", where there is an int or bool that is originally set to true and will turn to false once it is not a palindrome. This means after the loop you check if it is true or false and print the message accordingly. Here, we avoid the early return.

How do we make this a loop? Well we can start with int i = 0. So bringing what we had above, we could get:

int i = 0;
while (i < len) {
    // Remember to turn each letter to lowercase as mentioned in the original problem
    if (tolower(str[i]) != tolower(str[len - i])) {
        printf("String is not a palindrome\n");
        // early return if not a palindrome
        return 0;
    }
    i++;
}

Will this work? Looking closely at the if statement, we get

if (tolower(str[0]) != tolower(str[len - 0]))

which is not good because we wanted len - 1.

This just means that we should have str[len - i - 1] instead and if we try it for i = 1,2,3,... we see that it ensures these i values work too! So the final while loop looks like:

int i = 0;
while (i < len) {
    // Remember to turn each letter to lowercase as mentioned in the original problem
    if (tolower(str[i]) != tolower(str[len - i - 1])) {
        printf("String is not a palindrome\n");
        // early return if not a palindrome
        return 0;
    }
    i++;
}

If we find that it is not a palindrome, we have early returned so will no execute any code after the while loop. So after the while loop, we know that it did not early return and hence is definitely a palindrome. So right after the palindrome we can print the message that the word is a palindrome then return 0.