Data-Structure-and-Algorithm-Java-interview-kit
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Valid Palindrome
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;
can you check it and merge it @NirmalSilwal ?
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?
@NirmalSilwal , I tried but I couldn't find optimized solution. I think it's easy to understandable for beginners like me.
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;
}
}