String Interview Recipes

Introduction

Many string problems in coding interviews share the exact same foundational logic. Instead of re-inventing the wheel during an interview, you should memorize these standard "recipes" or templates.

Recipe 1: Checking for a Palindrome

A palindrome reads the same forwards and backwards.

Pythonic Way (Using Slicing)

If space complexity is not strictly constrained, slicing is the fastest and most readable way to check for a palindrome.

def is_palindrome_pythonic(s):
    return s == s[::-1]

- Time: \(O(N)\) - Space: \(O(N)\) (creates a reversed copy)

Optimal Way (Two Pointers)

If you must use \(O(1)\) space, use two pointers meeting in the middle.

def is_palindrome_optimal(s):
    left, right = 0, len(s) - 1

    while left < right:
        if s[left] != s[right]:
            return False
        left += 1
        right -= 1

    return True

- Time: \(O(N)\) - Space: \(O(1)\)

Recipe 2: Frequency Counting (Anagrams)

Many problems require you to count character frequencies (e.g., Valid Anagram, Longest Palindrome).

Pythonic Way (Using `Counter`)

collections.Counter is incredibly powerful and handles frequency counting automatically.

from collections import Counter

def is_anagram(s, t):
    return Counter(s) == Counter(t)

- Time: \(O(N)\) - Space: \(O(U)\) where \(U\) is the number of unique characters.

Array Way (Lowercase ASCII only)

If the problem specifies "lowercase English letters only," using an array of size 26 is technically more efficient and often preferred by strict interviewers.

def is_anagram_array(s, t):
    if len(s) != len(t):
        return False

    counts = [0] * 26

    for i in range(len(s)):
        counts[ord(s[i]) - ord('a')] += 1
        counts[ord(t[i]) - ord('a')] -= 1

    for count in counts:
        if count != 0:
            return False

    return True

- Time: \(O(N)\) - Space: \(O(1)\) (array is always size 26)

Recipe 3: Reversing Words in a Sentence

Given a sentence "hello world", reverse it to "world hello".

Pythonic Way

Leverage split() and join().

def reverse_words(s):
    words = s.split()
    return " ".join(reversed(words))

- Time: \(O(N)\) - Space: \(O(N)\)

Recipe 4: Removing Unwanted Characters

Given a string with punctuation and mixed casing, clean it for processing.

Using a List Comprehension

def clean_string(s):
    return "".join([char.lower() for char in s if char.isalnum()])