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Algorithms & data structures

Classic patterns — search, sort, stacks, queues, and practical complexity-aware code.

4 matches
Algorithms & data structures medium

How to Decode a String with Repeated Brackets in Python

Decodes strings with patterns like '3[a]2[bc]' by using a stack to handle nested and repeated bracket groups.

stack string-decoding algorithms
Python
def decode_string(s: str) -> str:
    stack = []
    current_num = 0
    current_str = ""

    for ch in s:
        if ch.isdigit():
            current_num = current_num * 10 + int(ch)
        elif ch == "[":
            stack.append((current_str, current_num))
            current_str = ""
            current_num = 0…
13 0 Open
Algorithms & data structures easy

How to Map Strings to Uppercase in Python

Loops through a list of strings and builds a new list with each string converted to uppercase.

string loop uppercase
Python
strings = ["hello", "world", "python", "skillset"]

uppercased = []
for s in strings:
    uppercased.append(s.upper())

print(uppercased)
15 0 Open
Algorithms & data structures easy

How to compress consecutive numbers into range strings in Python

Convert a sorted list of consecutive integers into compact range strings like '1-3', '5-6', and '15'.

ranges compression arrays
Python
def compress_ranges(nums):
    """Convert a list of sorted consecutive numbers into range strings."""
    if not nums:
        return []
    
    ranges = []
    start = prev = nums[0]
    
    for num in nums[1:]:
        if num == prev + 1:
            prev = num
        else:
            if start == prev:
         …
14 0 Open
Algorithms & data structures easy

Split a String into Multiple Lines by Width in Python

Demonstrates a word-wrap algorithm that splits a message into rows without exceeding a maximum width.

strings word-wrap algorithm
Python
def split_message(text, max_width):
    words = text.split()
    rows = []
    current_row = []

    for word in words:
        if len(" ".join(current_row + [word])) > max_width:
            rows.append(" ".join(current_row))
            current_row = [word]
        else:
            current_row.append(word)

    if …
14 0 Open

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