Strings¶

Master string manipulation and pattern matching techniques.

Overview¶

Strings are immutable sequences of characters in Python.

# Creation
s1 = "Hello"
s2 = 'World'
s3 = """Multi-line
string"""

# Escape characters
escaped = "Line 1\nLine 2\tTabbed"

# Raw strings
raw = r"C:\Users\name"  # Backslashes preserved

Common Operations¶

s = "Hello, World!"

# Case
s.lower()           # "hello, world!"
s.upper()           # "HELLO, WORLD!"
s.title()           # "Hello, World!"
s.capitalize()      # "Hello, world!"

# Search
s.find("World")     # 7 (index) or -1
s.index("World")    # 7 or ValueError
s.count("l")        # 3

# Modify
s.replace("World", "Python")
s.strip()           # Remove whitespace
s.split(", ")       # ["Hello", "World!"]
", ".join(["a", "b"])  # "a, b"

# Check
s.startswith("Hello")  # True
s.endswith("!")        # True
"World" in s           # True
s.isalpha()            # False (has punctuation)
s.isdigit()            # False

String Slicing¶

s = "Hello, World!"

s[0]        # "H"
s[-1]       # "!"
s[0:5]      # "Hello"
s[:5]       # "Hello"
s[7:]       # "World!"
s[::-1]     # "!dlroW ,olleH" (reversed)
s[::2]      # "Hlo ol!" (every 2nd char)

String Building¶

# Bad: String concatenation in loop
result = ""
for i in range(1000):
    result += str(i)  # O(n²) due to immutability

# Good: Use join
result = "".join(str(i) for i in range(1000))  # O(n)

# Good: Use f-strings
name = "Alice"
age = 30
message = f"{name} is {age} years old"

Pattern Matching¶

import re

text = "Contact: email@example.com"

# Search
match = re.search(r'\w+@\w+\.\w+', text)
if match:
    print(match.group())  # email@example.com

# Find all
emails = re.findall(r'[\w.]+@[\w.]+', text)

# Split
parts = re.split(r'\s+', "Hello   World  Python")

# Replace
cleaned = re.sub(r'\s+', ' ', "Hello   World")

Common Algorithms¶

Reverse String¶

def reverse_string(s: str) -> str:
    return s[::-1]

Palindrome Check¶

def is_palindrome(s: str) -> bool:
    cleaned = ''.join(c.lower() for c in s if c.isalnum())
    return cleaned == cleaned[::-1]

Anagram Check¶

from collections import Counter

def is_anagram(s1: str, s2: str) -> bool:
    return Counter(s1.lower()) == Counter(s2.lower())

Time Complexity¶

Operation	Time
Access	O(1)
Search	O(n)
Concatenate	O(n + m)
Slice	O(k)
Split	O(n)

Practice Files¶

build/data-structures/02-strings/string_basics.py
build/data-structures/02-strings/pattern_matching.py
build/challenges/leetcode-easy/0344_reverse_string.py

Next Topic¶

Continue to Hash Tables.