Coding Challenges: String Manipulation

String manipulation problems are among the most frequently asked coding challenges in early interview rounds — they test comfort with indexing, slicing, and iteration without requiring advanced algorithmic knowledge, making them the natural starting point for structured practice.

Challenge 1: Check If a String Is a Palindrome

def is_palindrome(text):
    cleaned = text.lower().replace(" ", "")
    return cleaned == cleaned[::-1]

print(is_palindrome("Madam"))         # True
print(is_palindrome("racecar"))       # True
print(is_palindrome("hello"))         # False
print(is_palindrome("nurses run"))    # True — ignoring spaces

The cleanest Python solution compares the cleaned string against its own reverse using slicing ([::-1]), avoiding any manual loop entirely.

Challenge 2: Reverse a String Without Using [::-1]

def reverse_string(text):
    result = ""
    for char in text:
        result = char + result   # prepend each character
    return result

print(reverse_string("Python"))   # nohtyP

Interviewers sometimes specifically forbid the slice shortcut to see whether you understand the underlying logic, not just Python's convenient syntax — be ready with both approaches.

Challenge 3: Count the Frequency of Each Character

def char_frequency(text):
    freq = {}
    for char in text:
        freq[char] = freq.get(char, 0) + 1
    return freq

print(char_frequency("banana"))
# {'b': 1, 'a': 3, 'n': 2}

The Same Problem, Using collections.Counter

from collections import Counter

print(Counter("banana"))
# Counter({'a': 3, 'n': 2, 'b': 1})

print(Counter("banana").most_common(1))
# [('a', 3)] — the most frequent character, with its count

Challenge 4: Check If Two Strings Are Anagrams

def is_anagram(a, b):
    return sorted(a.lower()) == sorted(b.lower())

print(is_anagram("listen", "silent"))   # True
print(is_anagram("hello", "world"))     # False

Sorting both strings and comparing the results is the simplest correct approach — two strings are anagrams if and only if they contain exactly the same characters in the same quantities, which sorting will always reveal.

Challenge 5: Count Vowels and Consonants

def count_vowels_consonants(text):
    vowels = "aeiouAEIOU"
    vowel_count = sum(1 for char in text if char in vowels)
    consonant_count = sum(1 for char in text if char.isalpha() and char not in vowels)
    return vowel_count, consonant_count

v, c = count_vowels_consonants("Hello World")
print(f"Vowels: {v}, Consonants: {c}")   # Vowels: 3, Consonants: 7

Challenge 6: Find the First Non-Repeating Character

def first_unique_char(text):
    freq = {}
    for char in text:
        freq[char] = freq.get(char, 0) + 1

    for char in text:
        if freq[char] == 1:
            return char
    return None

print(first_unique_char("swiss"))   # w

The first loop builds a frequency map; the second loop walks through the string again in its original order, returning the first character whose count is exactly 1.

Time Complexity Summary

Interview tip: For nearly every string problem, interviewers want you to state the time complexity of your solution out loud without being asked — even a correct answer feels incomplete to many interviewers if you cannot explain why it runs in O(n) or O(n log n) time.