Strings

What you'll learn

How strings behave as arrays of characters, and how to solve classic problems: reversing, palindromes and anagrams, using counting and two pointers.

Why it matters

Text is everywhere: search boxes, usernames, URLs, logs. String problems also appear in nearly every coding interview, and they reuse tools you already know: arrays, hash maps and loops.

Explanation

A string is a sequence of characters. You can read by index (s[0]) and ask its length, but in JavaScript strings are immutable: you cannot change a character in place. Every "change" creates a new string.

Useful tools:

• s.split("") turns a string into an array of characters. arr.join("") turns it back.
• s.toLowerCase(), s.slice(a, b), s.includes(x), s.indexOf(x).
• Loop with for (const ch of s).

Three patterns to know:

1. Counting with a hash map: count each character. Two words are anagrams (the same letters in a different order) if their counts match.
2. Two pointers: one index at the start and one at the end, moving toward each other. A palindrome reads the same both ways, so compare s[left] with s[right].
3. Building results in an array, then join at the end, because repeated string concatenation in a loop can be slow in some languages.

Complexity: most of these are O(n), one pass over the text.

Example

Step-by-step walkthrough

1. reverse splits "hello" into letters, reverses the array and joins it: olleh.
2. isPalindrome removes spaces, punctuation and case, giving amanaplanacanalpanama.
3. Two pointers start at both ends and compare letters while moving inward. They all match, so it prints true. For "hello", h and o differ, so it prints false.
4. isAnagram counts the letters of the first word, then subtracts for each letter of the second. If any count is missing, they are not anagrams. listen and silent use the same letters: true. abc and abd do not: false.
5. All three do one pass, so they are O(n).

Try it yourself

Write countVowels(s) that returns the number of vowels in a string, then test it on "Data Structures".

String problemsJavaScript

Editable codefunction isPalindrome(s) { const clean = s.toLowerCase().replace(/[^a-z0-9]/g, ""); let left = 0; let right = clean.length - 1; while (left < right) { if (clean[left] !== clean[right]) return false; left++; right--; } return true; } console.log(isPalindrome("Racecar")); console.log(isPalindrome("Not a palindrome"));

Run ResetCopyCtrl+Enter to run

Common mistakes

• Trying to change a character with s[0] = "x". Strings are immutable.
• Forgetting case and punctuation in palindrome checks.
• Comparing sorted strings when counting would be faster (sorting is O(n log n)).
• Off-by-one in two pointers. Use left < right, not <=.

Quick quiz

Summary

• Strings are indexable and immutable; convert to arrays to rearrange.
• Counting with a hash map solves anagrams; two pointers solve palindromes.
• Most string tasks are a single O(n) pass.

Next: recursion, functions that call themselves.

function reverse(s) {
  return s.split("").reverse().join("");
}

function isPalindrome(s) {
  const clean = s.toLowerCase().replace(/[^a-z0-9]/g, "");
  let left = 0;
  let right = clean.length - 1;
  while (left < right) {
    if (clean[left] !== clean[right]) return false;
    left++;
    right--;
  }
  return true;
}

function isAnagram(a, b) {
  if (a.length !== b.length) return false;
  const counts = new Map();
  for (const ch of a) counts.set(ch, (counts.get(ch) || 0) + 1);
  for (const ch of b) {
    if (!counts.get(ch)) return false;
    counts.set(ch, counts.get(ch) - 1);
  }
  return true;
}

console.log(reverse("hello"));
console.log(isPalindrome("A man, a plan, a canal: Panama"));
console.log(isPalindrome("hello"));
console.log(isAnagram("listen", "silent"));
console.log(isAnagram("abc", "abd"));