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Rank Combined Titles by Length, Return Top N — Automation For Makers

Day 7 · ~12 min●

You can aggregate and filter. Now rank. Combine text from three APIs, sort by character length from longest to shortest, return the top N. What Python function does the sort?

sorted() with a key argument? But I always forget how reverse works.

sorted(items, key=len, reverse=True) gives you longest-first. key=len tells sort to compare by length, reverse=True flips ascending to descending:

python

ranked = sorted(combined, key=len, reverse=True)
top = ranked[:n]

Slicing ranked[:n] handles over-indexing safely — asking for 100 from a 3-item list just returns all 3?

Exactly. Python slice semantics mean you never hit an IndexError. Here's the full function with three reads, one sort, one slice:

python

def rank_by_length(n: int, max_items: int) -> list:
    emails = toolset.execute_action(Action.GMAIL_FETCH_EMAILS, {"max_results": max_items})
    events = toolset.execute_action(Action.GOOGLECALENDAR_FIND_EVENT, {"query": ""})
    tasks = toolset.execute_action(Action.GOOGLETASKS_LIST_TASKS, {"max_results": max_items})
    combined = (
        [m.get("snippet", "") for m in emails.get("messages", [])]
        + [e.get("summary", "") for e in events.get("items", [])]
        + [t.get("title", "") for t in tasks.get("items", [])]
    )
    ranked = sorted(combined, key=len, reverse=True)
    top = ranked[:n]
    print(f"Top {len(top)} items by length from {len(combined)} total")
    return top

Is sorted() stable? If two items have the same length, is the original order preserved?

Yes. Python's sort is stable — equal keys preserve their relative order. That means items from Gmail keep their relative order among themselves even after the sort, and the same for events and tasks.

So one call gives me the longest pieces of text across my entire inbox, calendar, and task list — the stuff most likely to have real content?

Three live APIs, one sort, one slice. Top of the list is always the densest content. That closes Week 1 — next week you start shaping the richer structures this ranking feeds into.

Sort + Slice

TL;DR: sorted(items, key=len, reverse=True)[:n] gives you the top N by length across any list of strings — one line, stable, safe on small lists.

key=len — sort by length, not alphabetically
reverse=True — longest first
Slice [:n] — safe even when n exceeds len(items); returns the whole list if fewer than n exist
Stable — items with equal length keep their original relative order

Stable sort guarantee

Input	Output
two items with same `len`	original order preserved
two items with different `len`	longer one first
`n > len(items)`	whole list, no error

Day 7 · ~12 min●

You can aggregate and filter. Now rank. Combine text from three APIs, sort by character length from longest to shortest, return the top N. What Python function does the sort?

sorted() with a key argument? But I always forget how reverse works.

sorted(items, key=len, reverse=True) gives you longest-first. key=len tells sort to compare by length, reverse=True flips ascending to descending:

python

ranked = sorted(combined, key=len, reverse=True)
top = ranked[:n]

Slicing ranked[:n] handles over-indexing safely — asking for 100 from a 3-item list just returns all 3?

Exactly. Python slice semantics mean you never hit an IndexError. Here's the full function with three reads, one sort, one slice:

python

def rank_by_length(n: int, max_items: int) -> list:
    emails = toolset.execute_action(Action.GMAIL_FETCH_EMAILS, {"max_results": max_items})
    events = toolset.execute_action(Action.GOOGLECALENDAR_FIND_EVENT, {"query": ""})
    tasks = toolset.execute_action(Action.GOOGLETASKS_LIST_TASKS, {"max_results": max_items})
    combined = (
        [m.get("snippet", "") for m in emails.get("messages", [])]
        + [e.get("summary", "") for e in events.get("items", [])]
        + [t.get("title", "") for t in tasks.get("items", [])]
    )
    ranked = sorted(combined, key=len, reverse=True)
    top = ranked[:n]
    print(f"Top {len(top)} items by length from {len(combined)} total")
    return top

Is sorted() stable? If two items have the same length, is the original order preserved?

So one call gives me the longest pieces of text across my entire inbox, calendar, and task list — the stuff most likely to have real content?

Three live APIs, one sort, one slice. Top of the list is always the densest content. That closes Week 1 — next week you start shaping the richer structures this ranking feeds into.

Sort + Slice

TL;DR: sorted(items, key=len, reverse=True)[:n] gives you the top N by length across any list of strings — one line, stable, safe on small lists.

key=len — sort by length, not alphabetically
reverse=True — longest first
Slice [:n] — safe even when n exceeds len(items); returns the whole list if fewer than n exist
Stable — items with equal length keep their original relative order

Stable sort guarantee

Input	Output
two items with same `len`	original order preserved
two items with different `len`	longer one first
`n > len(items)`	whole list, no error

def rank_by_length(n: int, max_items: int) -> list: emails = toolset.execute_action(Action.GMAIL_FETCH_EMAILS, {"max_results": max_items}) events = toolset.execute_action(Action.GOOGLECALENDAR_FIND_EVENT, {"query": ""}) tasks = toolset.execute_action(Action.GOOGLETASKS_LIST_TASKS, {"max_results": max_items}) combined = ( [m.get("snippet", "") for m in emails.get("messages", [])] + [e.get("summary", "") for e in events.get("items", [])] + [t.get("title", "") for t in tasks.get("items", [])] ) ranked = sorted(combined, key=len, reverse=True) top = ranked[:n] print(f"Top {len(top)} items by length from {len(combined)} total") return top

Sort + Slice

TL;DR: sorted(items, key=len, reverse=True)[:n] gives you the top N by length across any list of strings — one line, stable, safe on small lists.

key=len — sort by length, not alphabetically

reverse=True — longest first

Slice [:n] — safe even when n exceeds len(items); returns the whole list if fewer than n exist

Stable — items with equal length keep their original relative order

Stable sort guarantee

Input	Output
two items with same `len`	original order preserved
two items with different `len`	longer one first
`n > len(items)`	whole list, no error

Sort + Slice

TL;DR: sorted(items, key=len, reverse=True)[:n] gives you the top N by length across any list of strings — one line, stable, safe on small lists.

key=len — sort by length, not alphabetically

reverse=True — longest first

Slice [:n] — safe even when n exceeds len(items); returns the whole list if fewer than n exist

Stable — items with equal length keep their original relative order

Stable sort guarantee

Input	Output
two items with same `len`	original order preserved
two items with different `len`	longer one first
`n > len(items)`	whole list, no error

Sort + Slice

Stable sort guarantee

Sort + Slice

Stable sort guarantee

Sort + Slice

Stable sort guarantee

Sign up to practice

Sort + Slice

Stable sort guarantee

Sign up to practice