Strings

Character counting, palindromes, and parsing.

Reach for it when

Palindromic substrings (expand around center)
Character frequency comparisons
Parse or build strings efficiently

Core idea

Strings are immutable in Python, so build results in a list and join once. For palindromes, expand around each of the 2n - 1 centers. For anagram-like questions, compare fixed-size frequency arrays of length 26.

Template · O(n²) for center expansion, O(n) for counting

def count_palindromes(s):
    count = 0
    for center in range(2 * len(s) - 1):
        l, r = center // 2, (center + 1) // 2
        while l >= 0 and r < len(s) and s[l] == s[r]:
            count += 1
            l, r = l - 1, r + 1
    return count

int countPalindromes(const string& s) {
    int n = s.size(), count = 0;
    for (int center = 0; center < 2 * n - 1; center++) {
        int l = center / 2, r = (center + 1) / 2;
        while (l >= 0 && r < n && s[l] == s[r]) {
            count++;
            l--; r++;
        }
    }
    return count;
}

Common pitfalls

Repeated string concatenation in a loop is O(n²)
Forgetting even-length centers

Strings

Reach for it when

Core idea

Template · O(n²) for center expansion, O(n) for counting

Common pitfalls

Practice · 0 / 6 done