String Manipulation String Hash Map Counting Character Pattern Matching
Intermediate Level

Implement the stringCompressionLength method that returns the length after compressing consecutive repeated characters.

The input contains a string text. Your task is to return the length of the compressed version of the string.

Compression is based on consecutive repeated characters. For each group, write the character followed by how many times it appears. If the compressed version is not shorter than the original string, return the original string length instead.

For example, aabbccc becomes a2b2c3, so the compressed length is 6. For abc, compression would not reduce the string, so the answer remains 3.

Example 1
Input:
text (string) = aaabbc
Return:
(int) 6
Example 2
Input:
text (string) = abc
Return:
(int) 3
Example 3
Input:
text (string) = aabbccc
Return:
(int) 6

Scan the string from left to right and count consecutive occurrences of the same character.

Whenever the current group ends, add the length of the compressed group to a running total. A compressed group contains the character and its count, so aaa contributes the length of a3.

After calculating the compressed length, compare it with the original length. Return the smaller useful length so that the compression result never becomes longer than the original string.

Pseudocode:

function stringCompressionLength(text):
    if text is empty:
        return 0
    compressedLength = 0
    count = 1
    for i from 1 to length(text) - 1:
        if text[i] == text[i - 1]:
            count++
        else:
            compressedLength = compressedLength + 1 + numberOfDigits(count)
            count = 1
    compressedLength = compressedLength + 1 + numberOfDigits(count)
    return min(length(text), compressedLength)
Run your code to see the result.