Good morning! Here's our prompt for today.

Let's define the "compacting" of a string as taking an input like the following: 'sstttrrrr'.

We want to take the number of sequential appearances (number of times it shows up in a row) of each letter:

1s: 2
2t: 3
3r: 4

And denote them next to the original character in the string, getting rid of the duplicates. In our above example, our output is: 's2t3r4'.

How long is this compacted string? Could you write a method that takes in a string and returns the length of the compacted one? Could you solve it using only O(1) extra space?

Here are few other examples:

1compactLength('aabbbbbbbbbbbbb')
2// 5 because we end up with `a2b13`

Another to consider:

1compactLength('s')
2// 1 because we still end up with just `s`

Constraints

• Length of the given string <= 100000
• The string comprises of ASCII characters
• Expected time complexity : O(n)
• Expected space complexity : O(1)

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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​

###

# @param {character[]} str

# @return {number}

###

def compact_length(s):

    # fill this in

    return len(s)

​

​

# print(compactLength("a"))

# print(compactLength("abb"))

# print(compactLength("aabbbbbbbbbbbbb"))

​

​

import unittest

​

​

class Test(unittest.TestCase):

    def test_1(self):

        assert callable(compact_length) == True

        print("PASSED: 'compact_length' is a function")

​

    def test_2(self):

        assert compact_length("a") == 1

        print("PASSED: compact_length('a') should return 1")

​

    def test_3(self):

        assert compact_length("abb") == 3

        print("PASSED: compact_length('abb') should return 3")

We'll now take you through what you need to know.

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