First, I was thinking what if I use Counter from collections?
Then suddenly I remember the input and output. Counting the unique char is not the thing but counting the consecutive same character.
My initial logic is:
- I'll create a prev var to hold pre character.
- count to count current character
- and the output var
By looping from 1 to n-1 is my thing.
def compress(fullword):
l1 = len(fullword)
if l1<2:
return fullword
output = ""
prev = fullword[0]
count = 1
for s in fullword[1:]:
if s == prev:
count+=1
else:
output += (prev+str(count))
count = 1
prev = s
output += (prev+str(count))
return output if len(output) < len(fullword) else fullword
Space Complexity: O(len(word)) as looping though all the character.
Space Complexity: O(1) no extra variable is used except some constants.
test_cases = [
("aabcccccaaa", "a2b1c5a3"),
("abcdef", "abcdef"),
("aabb", "aabb"),
("aaa", "a3"),
("a", "a"),
("", ""),
]