查找字符串中每个字符出现的次数的复杂性

Question

input_string = "foobaarfoooobaaaarfo"
count_dict = {}
for char in input_string:
    try:
        count_dict[char]=count_dict[char]+1
    except:
        count_dict[char]=1
print(count_dict)

is this the best algorithm with complexity O(N) for the given problem statement

Answer 1

Yes, O(n) is the best you can do.

It is necessary to visit each character to count them all.

However, in terms of python implementation, you will probably get better performance, and more readable code using the specialized collection Counter :

from collections import Counter
input_string = "foobaarfoooobaaaarfo"
counter = Counter(input_string)
print(counter)

Answer 2

I agree that O(N) is the best you can do.

I did a simplification for your program. You do not have to use try and raise exception here.

input_string = "foobaarfoooobaaaarfo"
count_dict = {}
for char in input_string:
    count_dict[char] = count_dict.get(char, 0) + 1
print(count_dict)