String Compression

Problem statement : Write a function to Compress a given string. Only return the compressed string if it saves space.

Difficulty level: Easy

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Test Cases

• 'AAABAACCDDDD' ---> 'A3BA2C2D4'
• 'BAAACCDDDD' ---> 'BA3C2D4'
• 'AABBCC' --> 'AABBCC'
• 'BBBAACCCCDD' --> 'B3A2C4D2'

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Algorithm

• Setup a pointer at first character of string, initialise an empty string and a counter.
• Iterate through the string and check if character is equal to the first character of the string.
  • if both are equal increment counter.

• Else,

  • add the first character and it's count to result string.
    • if count is greater than 1 then only add count to the result string.
    • Else, add an empty string to the result string.
  • now, the first character will be the current character.
  • reset the counter value to 1.

• Add the last character and it's count to result string.

• Return the resultant compressed string if it's length is less than the given string, else return the given string.

Time and Space Complexity

• Time complexity: O(n)
• Space complexity: O(n)

---

Code

class CompressString(object):
    def compress(self, input):
        if input is None or not input:
            return input
        result = ''
        count = 0
        prev_char = input[0]
        for char in input:
            if char == prev_char:
                count += 1
            else:
                result += prev_char + (str(count) if count > 1 else '')
                prev_char = char
                count = 1
        result += prev_char + (str(count) if count > 1 else '')
        return result if len(result) < len(input) else input

More precise compression

class CompressString(object):
    def compress(self, input):
        if input is None or not input:
            return input
        result = ''
        count = 0
        prev_char = input[0]
        for char in input:
            if char == prev_char:
                count += 1
            else:
               if count > 2:
                    result += prev_char + (str(count))
                    prev_char = char
                    count = 1
               elif count == 1:
                    result += prev_char
                    count = 1
                    prev_char = char
               else:
                    result += prev_char
                    result += prev_char
                    count = 1
                    prev_char = char
        result += prev_char + (str(count) if count > 2 else prev_char)
        return result if len(result) < len(input) else input

Test

import unittest
from compressString import CompressString


class TestCompress(unittest.TestCase):

    def testCompress(self, func):
        self.assertEqual(func(None), None)
        self.assertEqual(func(''), '')
        self.assertEqual(func('AABBCC'), 'AABBCC')
        self.assertEqual(func('AAABCCDDDDE'), 'A3BC2D4E')
        self.assertEqual(func('BAAACCDDDD'), 'BA3C2D4')
        self.assertEqual(func('AAABAACCDDDD'), 'A3BA2C2D4')
        print('Success: testCompress')


def main():
    test = TestCompress()
    compress_string = CompressString()
    test.testCompress(compress_string.compress)


if __name__ == '__main__':
    main()

Original article

Github solution

Happy Coding !

Comments

[Ryszard Grodzicki]

Simple improvement proposal: you could remove the last check by replacing chars only if there are more than 2 in a row.
Replacing 'CC' with 'C2' never saves space. Replacing more than 2 characters always saves space. So you would either return the same input string or return compressed string that's shorter than original.

[CodeWithML]

I could do something like if count is more than 2 than add the character with count, if count is 1, than only add the character , and lastly if count is exactly 2, then add the character 2 times, so it would remain same as original string.

[CodeWithML]

I didn't know that, looks really neat. Thanks for the tip. 🌟