Suppose we have a list of words. These words may occur multiple times. We have to show the frequencies of these words and count how many distinct words are there.
So, if the input is like words = ["Book", "Sound", "Language", "Computer", "Book", "Language"], then the output will be (4, '2 1 2 1') because there are four distinct words, the first and third words have occurred twice.
To solve this, we will follow these steps −
- d:= an OrderedDict to store items based on insert order
- for each w in words, do
- if w is in d, then
- d[w] := d[w] + 1
- otherwise,
- d[w] := 1
- if w is in d, then
- a pair of the size of list of all keys in d and join all values from d into a string then return.
Example
Let us see the following implementation to get better understanding
from collections import OrderedDict def solve(words): d=OrderedDict() for w in words: if w in d: d[w] += 1 else: d[w] = 1 return len(d.keys()), ' '.join([str(d[k]) for k in d.keys()]) words = ["Book", "Sound", "Language", "Computer", "Book", "Language"] print(solve(words))
Input
["Book", "Sound", "Language", "Computer", "Book", "Language"]
Output
(4, '2 1 2 1')