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LESSON 122 OF 673 · LISTS - CHALLENGES

Add If Missing in Python: not in and append

Practice Read the full explanation

You get a shopping list items, a list of strings, and a word item. Add item at the end only when it isn't in the list yet. Return the list.

When item is already there, the list stays as it is. The list may start empty.

 

Example 1:

Input: items = ["milk", "eggs"], item = "bread"
Output: ["milk", "eggs", "bread"]
Explanation: The word "bread" is not in the list yet.
It goes at the end.

Example 2:

Input: items = ["milk", "eggs"], item = "milk"
Output: ["milk", "eggs"]
Explanation: The word "milk" is already there.
Nothing is added.

Example 3:

Input: items = [], item = "apples"
Output: ["apples"]
Explanation: The list is empty, so "apples" is new.
It becomes the only item.

 

Constraints:

  • items has 0 to 9 words: it may be empty.
  • item and every word in items are made of lowercase English letters.
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    PYTHON · LISTS - CHALLENGES

    Add If Missing in Python, explained

    Problem overview

    The method addIfMissing gets a shopping list items (a list of strings) and a word item. It adds item at the end only when the word isn't in the list yet, and returns the list. When the word is already there, the list stays as it is. The list may start empty.

    Examples:

    • ["milk", "eggs"] with "bread" gives ["milk", "eggs", "bread"].
    • ["milk", "eggs"] with "milk" gives ["milk", "eggs"]. The word is already there.
    • [] with "apples" gives ["apples"].

    Approach: ask first, then add

    This is a question followed by an action. The question is "is the word missing?", which item not in items answers with True or False. The operator checks the list item by item and compares values. The action is items.append(item), which adds at the end. Only when the answer is True does the append run.

    The empty list needs no special case. Nothing is in an empty list, so item not in items is True and the word is appended, which makes it the only item.

    Step by step

    1. Write an if whose condition is item not in items.
    2. Append item inside the if. When the word is already in the list, the body is skipped and nothing changes.
    3. Return items after the if, so the line runs in both cases.

    The solution in Python

    class Solution:
    def addIfMissing(self, items: List[str], item: str) -> List[str]:
        if item not in items:
            items.append(item)
        return items

    On the three examples it prints:

    ['milk', 'eggs', 'bread']
['milk', 'eggs']
['apples']

    The return is indented at the level of the if, not inside it. Put inside the if, it would return nothing (None) whenever the word was already there. Also remember that append returns None: write items.append(item) on its own line, and never items = items.append(item).

    Another way: return early

    You can write the check the other way round and leave the function as soon as the word is found:

    def add_if_missing(items, item):
    if item in items:
        return items
    return items + [item]

    It gives ['milk', 'eggs', 'bread'], ['milk', 'eggs'] and ['apples'] on the three examples. Here the last line builds a new list with items + [item], so the list passed in is not changed. The reference solution is the shorter of the two and keeps one return at the bottom.

    Edge cases and common mistakes

    • Putting the return inside the if. When the word is already in the list, the function then returns None: the call addIfMissing(["milk", "eggs"], "milk") would print None.
    • Appending without the check. ["milk", "eggs"] with "milk" would become ['milk', 'eggs', 'milk'].
    • Using in where not in is needed. Then the word is added only when it is already there, which is the opposite of the task.
    • Text must match exactly. "Milk" is a different word from "milk". The constraints use lowercase letters only, so this does not come up in the tests.
    • Making a new list. items + [item] builds a new list and leaves items alone. For the first example it gives the right list, but it is a different one, and the original list is not changed. The reference solution changes the given list instead.

    Testing

    Try a word that is the first item of the list, one that is the last item, and the empty list. The single not in test handles all three the same way. Printing the result is enough to see the difference between an added word and an unchanged list.

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