How to compress a string in-place - arrays

Question

I have been working on this problem on leetcode https://leetcode.com/problems/string-compression/
Given an array of characters, compress it in-place.
The length after compression must always be smaller than or equal to the original array.
Every element of the array should be a character (not int) of length 1.
After you are done modifying the input array in-place, return the new length of the array.
I almost have a solution, but I can't seem to count the last character in the string and I also am not sure how to make it so if there is only an amount of one of a character that I do not show 1 in the array.
I feel like I'm pretty close and I'd like to try and keep the solution that I have without altering it too much if possible.
This is what I have so far. chars is a list of characters
def compress(chars):
char = 0
curr = 0
count = 0
while curr < len(chars):
if chars[char] == chars[curr]:
count += 1
else:
# if count == 1:
# break
# else:
chars[char-1] = count
char = curr
count = 0
curr += 1
chars[char-1] += 1
return chars
print(compress(["a", "a", "b", "b", "c", "c", "c"]))

Answer 1

I wasn't quite able to format your code to get the answer you were seeking. Based on your answer, I was able to put together code and explanation that could help you:
def compress(chars):
count = 1
current_position = 0
# if it's a single character, just return a
# a basic array with count
if len(chars) == 1:
chars.append("1")
return chars
# loop till the 2nd last character is analyzed
while current_position < len(chars) - 1:
# assume that we haven't reached the 2nd last character
# if next character is the same as the current one, delete
# the current one and increase our count
while current_position < len(chars) - 1 and \
chars[current_position] == chars[current_position + 1]:
del chars[current_position]
count += 1
# if next character isn't the same, time to add the count to
# the list. Split the number into
# character list (e.g. 12 will become ["1", "2"]
# insert those numbers behind the character and increment position
for x in str(count):
chars.insert(current_position + 1, str(x))
current_position += 1
# great, on to the next character
current_position += 1
# if we are now at the last character, it's a lonely character
# give it a counter of 1 and exit the looping
if current_position == len(chars) - 1:
chars.append("1")
break
count = 1
return chars
mylist = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
print(compress(mylist))
Results
mylist = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
['a', '1', 'b', '1', '2']
mylist = ["a","a","a","a","a","a","a","a","a","a","b","b","b","b","b","b","b","b","b","b","b","b"]
['a', '1', '0', 'b', '1', '2']
mylist = ["a"]
['a', '1']
mylist = ["a","b"]
['a', '1', 'b', '1']
mylist = ["a","a","b","b","c","c","c"]
['a', '2', 'b', '2', 'c', '3']