How to check if word is palindrome using substrings instead of charAt?

Question

I just realized my mistake.

Instead of writing

test1 = do.substring(x, x + 1);
test2 = do.substring(count - 1 - x, count - x);

I wrote

test1 = do.substring(x, x + 1);
test1 = do.substring(count - 1 - x);

I was not aware that you had to "end" a substring, I used this answer. Considering the upvotes it has, I was wondering if someone could explain why it didn't work in my case. How do I get the last character of a string?

Update: I made it so that the loop would print match/nomatch for each character. I see that the first loop matches, but for the 2nd and 3rd characters there is no match. I used the input "pop". — om242515, Nov 18 '20 at 17:43
This is what I got : pop match. nomatch. nomatch. This is not a palindrome. — om242515, Nov 18 '20 at 17:44
Not sure if you're allowed to use it, but you could split the input in half and then [use StringBuilder to reverse it](https://stackoverflow.com/questions/7569335/reverse-a-string-in-java) and see if they're equal. Just need to be mindful of odd length Strings when splitting. — Tim Hunter, Nov 18 '20 at 17:46
I will definitely try to learn more about stringbuilder for personal use, but this is sadly a HS compsci course that's taught for a single exam — om242515, Nov 18 '20 at 17:48
You able to use [toCharArray](https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/lang/String.html#toCharArray())? — Tim Hunter, Nov 18 '20 at 17:49
Unfortunately I can't, that was another way i was planning to write the program before i was told we could only use substrings — om242515, Nov 18 '20 at 17:52
Just realized my mistake! I didn't define (?) the substring correctly. — om242515, Nov 18 '20 at 17:58
The second parameter tells the `substring` method what the ending index should be. Without it, the method assumes you want a substring ending at the far end of the string. — Tim Hunter, Nov 18 '20 at 17:58

Mustafa Poya · Answer 1 · 2020-11-18T18:33:54.913

try to convert your string to an charArray and then define to indexed one from start of array other from end of array and compare them with each other: if they are always the same then its a Palindrome else its not.

String input = "POP";
char [] values = input.toCharArray();
boolean isPalindrome = true;

for (int i = 0, j = values.length-1; i < values.length /2 && j >= values.length/2; i++, j--) {
    if(values[i] != values[j]) {
        isPalindrome = false;
        break;
    }
}
if(isPalindrome) {
    System.out.println(input + " is a Palindrom");
} else {
    System.out.println(input + " is not a Palindrom");
}

score 0 · Answer 2 · answered Nov 18 '20 at 17:57

I think you don't need to iterate over string when using substrings. You can simply compare left substring with reversed right substring like this:

private static boolean isPalindrome(final String input) {
    int length = input.length();
    int middleIndex = (length % 2 == 0 ) ?
            (length / 2 ) : (length / 2 + 1);
    String left = input.substring(0, length / 2);
    String right = input.substring(middleIndex, length);

    // Compare left string with reverse(right)
    return left.equals(new StringBuilder(right).reverse().toString());
}

Arvind Kumar Avinash · Answer 3 · 2021-05-20T05:28:00.780

Quoted below is the description of String#substring(int beginIndex, int endIndex):

Returns a new string that is a substring of this string. The substring begins at the specified beginIndex and extends to the character at index endIndex - 1. Thus the length of the substring is endIndex-beginIndex.

Thus, in order to get a string of just one character using String#substring(int beginIndex, int endIndex), the value of endIndex needs to be equal to beginIndex + 1 e.g. "Hello".substring(1, 2) is equivalent to String.valueOf("Hello".charAt(1)).

Demo:

public class Main {
    public static void main(String[] args) {
        // Test strings
        String[] arr = { "dad", "malayalam", "papa" };
        for (String s : arr) {
            System.out.println(isPallindrome(s));
        }
    }

    static boolean isPallindrome(String s) {
        int len = s.length();
        for (int i = 0; i < len / 2; i++) {
            if (!s.substring(i, i + 1).equals(s.substring(len - i - 1, len - i))) {
                return false;
            }
        }
        return true;
    }
}

Output:

true
true
false

How to check if word is palindrome using substrings instead of charAt?

3 Answers3