Two pointers on strings

Two-pointer problems are the exam's String FRQ heart. Model: one index from the front, one from the back, meeting in the middle.

"Is s a palindrome, considering letters only?" — the filter + two pointers combined:

public static boolean isPal(String s) {
    int lo = 0, hi = s.length() - 1;
    while (lo < hi) {
        if (!Character.isLetter(s.charAt(lo))) { lo++; }
        else if (!Character.isLetter(s.charAt(hi))) { hi--; }
        else if (s.charAt(lo) != s.charAt(hi)) { return false; }
        else { lo++; hi--; }
    }
    return true;
}

Trace "a, b a" (a comma, a space, letters a b a): lo=0 'a' vs hi=5 'a' match; lo=1 ',' skipped (lo++); lo=2 ' ' skipped; lo=3 'b' vs hi=4 'b' match; lo=4, hi=3 → loop ends, true. Notice the loop condition lo < hi — crossed pointers mean success, and single-character strings are trivially palindromes (0 < 0 is false immediately).

The second classic: run-length counting — count runs of equal adjacent characters:

public static int runs(String s) {
    if (s.length() == 0) { return 0; }
    int count = 1;
    for (int i = 1; i < s.length(); i++) {
        if (s.charAt(i) != s.charAt(i - 1)) {
            count++;
        }
    }
    return count;
}

"aabbbca" → runs a|bb|c|a = 4. Start at 1 (the first character always opens a run); increment at every boundary. The precondition length >= 1 is what lets you seed count = 1 — remove it and the empty string breaks you.