Lesson 4 / 42

Strings

Q: Fastest way to test if two strings are anagrams?

Compare character counts. Counting each character is O(n); permutations are O(n!).

An array of characters — often immutable, which changes how you build them.

Immutability matters

In Python, Java, and JS a string can't be changed in place. s += c in a loop builds a brand-new string each time — O(n^2) overall. Collect chars in a list/StringBuilder and join once.

Build efficiently

The list-and-join pattern is linear.

# slow: O(n^2)
out = ""
for ch in text:
    out += ch.upper()

# fast: O(n)
parts = []
for ch in text:
    parts.append(ch.upper())
out = "".join(parts)

Common toolkit

Frequency counts (hash map / array of 26), two pointers for palindromes, sliding window for substrings, and sorting characters as an anagram key.

Quick check: Fastest way to test if two strings are anagrams?

Try every permutation
Compare character counts
Reverse one and compare

Answer

Compare character counts — Counting each character is O(n); permutations are O(n!).