from itertools import groupby

input_value = input()

group_result = groupby(input_value)

for key,value in group_result:

value = sorted(list(value))
print(f"({int(len(value))}, {int(key)})", end=" ")

Core Python code (Without Built-in function)- We will appreciate itertools module some other time :

string= input()

lst= [element for element in string]

count= 1

prev= lst[0]

for i in range(1, len(lst)):
    if lst[i] == prev:
        count = count + 1

    else:
        print(f"({count}, {prev})", end= " ")
        prev= lst[i]
        count = 1

print(f"({count}, {prev})")

Here is HackerRank Compress the String! in python solution - https://programmingoneonone.com/hackerrank-compress-the-string-solution-in-python.html

import itertools

if name == 'main': S = input()

result = [(len(list(group)), int(key)) for key, group in itertools.groupby(S)]    


for item in result:
    print(item , end=' ')