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Valid Palindrome by ifenil · Pull Request #2 · NirmalSilwal/Data-Structure-and-Algorithm-Java-interview-kit · GitHub

Valid Palindrome - #2

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Valid Palindrome#2
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ifenil commented Oct 3, 2020
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Valid Palindeome From Leetcode August Challange ;
explanation of code line by line through comment;
easily understandable and simple code
it passed all the test cases;

ifenil commented Oct 3, 2020

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can you check it and merge it @NirmalSilwal ?

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Hi @ifenil, could you add another approach or more optimized solution. That would be better PR if you are considering hactoberfest.
What do you think?

ifenil commented Oct 3, 2020
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@NirmalSilwal ,
I tried but I couldn't find optimized solution.
I think it's easy to understandable for beginners like me.

javadev commented Mar 17, 2022

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The alternative solution

public class Solution {
    public boolean isPalindrome(String s) {
        int i = 0;
        int j = s.length() - 1;
        boolean res = true;
        while (res) {
            // Iterates through string to find first char which is alphanumeric.
            // Done to ignore non-alphanumeric characters.
            // Starts from 0 to j-1.
            while (i < j && isNotAlphaNumeric(s.charAt(i))) {
                i++;
            }
            // Similarly from j-1 to 0.
            while (i < j && isNotAlphaNumeric(s.charAt(j))) {
                j--;
            }
            // Checks if i is greater than or equal to j.
            // The main loop only needs to loop n / 2 times hence this condition (where n is string
            // length).
            if (i >= j) {
                break;
            }
            // Assigning found indices to variables.
            // The upperToLower function is used to convert characters, if upper case, to lower
            // case.
            // If already lower case, it'll return as it is.
            char left = upperToLower(s.charAt(i));
            char right = upperToLower(s.charAt(j));
            // If both variables are not same, result becomes false, and breaks out of the loop at
            // next iteration.
            if (left != right) {
                res = false;
            }
            i++;
            j--;
        }
        return res;
    }

    private boolean isNotAlphaNumeric(char c) {
        return (c < 'a' || c > 'z') && (c < 'A' || c > 'Z') && (c < '0' || c > '9');
    }

    private boolean isUpper(char c) {
        return c >= 'A' && c <= 'Z';
    }

    private char upperToLower(char c) {
        if (isUpper(c)) {
            c = (char) (c + 32);
        }
        return c;
    }
}

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Choose a reason Spam Abuse Off Topic Outdated Duplicate Resolved Low Quality

My solution to the problem. Missing some edge cases(which can be added later).

public class Palindrome {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    System.out.print("Please enter the string: ");
    String str = input.nextLine();
    input.close();

    int i = 0;
    int j = str.length() - 1;
    while (i < j) {
      if (str.charAt(i) == str.charAt(j)) {
        i++;
        j--;
      } else {
        System.out.printf("The string %s is not a Palindrome. \n", str);
        return;
      }
    }
    System.out.printf("The string %s is a Palindrome. \n", str);
  }
}

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