Discussion on Compress the String! Challenge

4 months ago+ 1 comment

Without itertools using sliding window concept.

from collections import defaultdict
strings = str(input())

dicts = defaultdict()
answers = []
left = 0
right = 1

if len(strings) == 1:
    print(f'({1}, {int(strings)})')

while(right < len(strings)):
    if strings[left] == strings[right]:
        if strings[left] not in dicts.keys():
            dicts[strings[left]] = 2
        else:
            dicts[strings[left]] += 1
        if right == len(strings) - 1:
            tuples = tuple([dicts[strings[left]],int(strings[left])])
            answers.append(tuples)
            dicts = defaultdict()
                
    else:
        if strings[left] not in dicts.keys():
            dicts[strings[left]] = 1
        tuples = tuple([dicts[strings[left]],int(strings[left])])
        answers.append(tuples)
        dicts = defaultdict()
        dicts[strings[right]] = 1
        left = right
        if right == len(strings) - 1:
            dicts[strings[left]] = 1
            tuples = tuple([dicts[strings[left]],int(strings[left])])
            answers.append(tuples)
            
    right += 1
    
		

    
print(' '.join(f'({x}, {y})' for x,y in answers))

2 weeks ago+ 0 comments

Wow. Incredibly long and complex. I wrote a solution without groupby(), too, but it was simple. Maybe the "sliding window" technique is not suited to this problem.

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