Session 6

Assignment

Complete get_last_index. Return the length of the inventory list minus 1.

Solution

def get_last_index(inventory):
    return len(inventory) - 1

Concept

len() function, last index = len(list) - 1

How it went

Solved first run, recognised the one-liner approach independently.

Ratings

Key takeaways

• →Simple one-liner patterns often work — read the problem twice before overcomplicating

Assignment

Complete smelt_ore. Check if inventory[1] == 'Iron Ore'. If so, change it to 'Iron Bar'. Return the inventory.

Solution

def smelt_ore(inventory):
    if inventory[1] == "Iron Ore":
        inventory[1] = "Iron Bar"
    return inventory

Concept

Mutating a list by index: list[i] = new_value

How it went

Solved correctly, one small lookup (index reassignment syntax) via search rather than asking — found the pattern and adapted it. Solution was clean and minimal.

Ratings

Key takeaways

• →Assigning to list[i] changes the original — not a copy. Worth remembering when lists get passed between functions

Assignment

Write a countdown from 10 to 1 printing '10...', '9...' etc. At 1 print '1...Fight!'

Solution

def countdown_to_start():
    for i in range(10, 0, -1):
        if i == 1:
            print(f"{i}...Fight!")
        else:
            print(f"{i}...")

Concept

range() with negative step, conditional print on final iteration

How it went

Assignment

Complete award_enchantments. Use a counter that resets every 3 quests. Only award enchantment when counter reaches 3. Strength = quest_number * 5.

Solution

def award_enchantments(start, end, step):
    counter = 0
    for quest_number in range(start, end, step):
        counter += 1
        if counter < 3:
            continue
        counter = 0
        enchantment_strength = quest_number * 5
        print(f"Enchantment of strength {enchantment_strength} awarded for completing {quest_number} quests!")

Concept

continue statement to skip loop iterations, counter-based pattern

How it went

Key takeaways

• →continue skips to next iteration immediately — useful for avoiding unnecessary work inside loops

Assignment

Complete calculate_experience_points. Each level requires level * 5 XP. Level 1 starts at 0 XP. Return total XP at the given level.

Solution

def calculate_experience_points(level):
    total_xp = 0
    for i in range(1, level):
        total_xp = total_xp + (i * 5)
    return total_xp

Concept

Accumulator pattern, running total inside a for loop

How it went

Key takeaways

• →Accumulator pattern: a variable updated inside a loop that carries state across iterations