String Problems

Reversal, anagrams, character counting, patterns you'll see everywhere

Strings are the most common data type in real-world programming. Log files, API responses, user input, config values, all strings. These 4 patterns cover 80% of string problems.

Pattern 1: Reversals[::-1], that's it. Python's slice with step -1 reverses any sequence.

String reversalpython

s = "Hello, World!"
print(s[::-1])   # !dlroW ,olleH

# Check palindrome
def is_palindrome(s):
    s = s.lower().replace(" ", "")  # normalize
    return s == s[::-1]

print(is_palindrome("racecar"))   # True
print(is_palindrome("A man a plan a canal Panama"))  # True

Pattern 2: Anagram checkTwo words are anagrams if they have the same characters in different order. Sort both and compare.

Anagram checkpython

def is_anagram(s1, s2):
    return sorted(s1.lower()) == sorted(s2.lower())

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

Pattern 3: Character countingUse a dictionary or collections.Counter.

Character frequencypython

def char_count(s):
    counts = {}
    for ch in s:
        counts[ch] = counts.get(ch, 0) + 1
    return counts

print(char_count("hello"))   # {'h': 1, 'e': 1, 'l': 2, 'o': 1}

# Most common character
s = "programming"
counts = char_count(s)
most_common = max(counts, key=counts.get)
print(f"Most common: '{most_common}' ({counts[most_common]}x)")

How would you approach this? Think first, then expand.

Practice Exercises

0/4 solved

Exercise 1 of 4easy

⏱ 00:00

Reverse Words

Reverse the ORDER of words in a sentence (not the characters).
Given: sentence = "Hello World Python"Expected output: Python World Hello

solution.py

1 / 4

Exercise 2 of 4easy

⏱ 00:00

Find All Vowels

Count how many vowels (a,e,i,o,u) are in the string.
Given: text = "Python Programming"Expected output: Vowels: 5

solution.py

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Exercise 3 of 4medium

⏱ 00:00

First Non-Repeated Character

Find the first character that appears only once.
Given: s = "swiss"Expected output: First unique: w
(s appears twice, so skip it. w appears once, that's the answer)

solution.py

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Exercise 4 of 4hard

⏱ 00:00

Compress a String

Compress consecutive repeated characters into char+count format.
Given: s = "aabbbccdddd"Expected output: a2b3c2d4

solution.py

4 / 4

Solve all 4 exercises to unlock completion

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