Counting with a dict
A dict finds a key without scanning, which makes it the natural tool for "how many of each?". The classic loop is:
counts = {}
for item in items:
counts[item] = counts.get(item, 0) + 1counts.get(item, 0) gives the current count, or 0 the first time an item shows up. collections.Counter(items) builds the same dict in one call.
Once you have counts, many questions become one more step: the biggest count is max(counts.values()), items seen more than once are the keys whose count is above 1, and the number of pairs inside a group of c equal items is c * (c - 1) // 2.
Watch the counts column in the table: one key grows by one on every pass.
See it run
words = ["tea", "jam", "tea", "bun", "tea"] counts = {} for w in words: counts[w] = counts.get(w, 0) + 1 print(counts)
A tally in a dict: each word becomes a key, its value is how often it appeared.
Practice counting with a dict
Exercise 1
You are sorting a pile of trading stickers. Each sticker is named by a string in stickers.
Return the number of copies of your most common sticker, as an integer.
Rules:
- Names are case-sensitive:
"Owl"and"owl"are different stickers. - An empty pile gives
0.
Constraints: 0 ≤ len(stickers) ≤ 10^5.
Example: solution(["owl", "fox", "owl"]) returns 2 because "owl" appears twice.
Exercise 2
A library scanner logs a card number every time a card is swiped. swipes is the list of card numbers in the order they were logged.
Return a list of the card numbers that were swiped more than once, sorted from smallest to largest, with each number listed once.
Rules:
- If no card was swiped more than once, return
[]. - Card numbers can be zero or negative.
Constraints: 0 ≤ len(swipes) ≤ 10^5.
Example: solution([4, 7, 4, 9, 7, 4]) returns [4, 7] because 4 was swiped three times and 7 twice.
Exercise 3
A coat-check desk gives every coat a colour code. tags is the list of codes in the order the coats arrived. Two coats are twins when they sit at positions i < j and have the same code.
Return how many twin pairs there are, as an integer.
Rules:
- Each pair of positions counts once, so three coats with the same code make 3 pairs.
- Fewer than two coats gives
0.
Constraints: 0 ≤ len(tags) ≤ 8000; codes are integers.
Example: solution([1, 2, 1, 1]) returns 3: positions (0,2), (0,3) and (2,3).
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Practice
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