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String Compression III - Step-by-Step Explanation

This repository provides a detailed explanation of how to solve the String Compression III problem across multiple programming languages. Here, you will find explanations for each language in a point-by-point format, guiding you through the problem-solving process step by step without revealing the full code.

Problem Overview

Given a string word, compress it by grouping contiguous occurrences of each character up to a maximum of 9. Each group is represented by the count followed by the character. For instance:

  • Input: word = "aaaaaaaaaaaaaabb"
  • Output: comp = "9a5a2b"

Steps for All Languages

The following explanations break down the steps needed to solve this problem in each of the five programming languages.

C++ Code

  1. Initialize a Result String and Counter:

    • Create an empty string, comp, to store the compressed result.
    • Set an integer variable, count, to 1 to begin counting the occurrences of each character.

  2. Iterate Over Each Character:

    • Loop through each character in word starting from the second character.
    • For each character, check if it:
      • Differs from the previous character.
      • Reaches a count of 9.
      • Is the last character in the string.
    • In any of these cases, append the count and the previous character to comp.

  3. Reset or Increment Counter:

    • If the above conditions are met, reset count to 1 to start counting the next character group.
    • Otherwise, increment count by 1 if the current character is the same as the previous one.

  4. Return Compressed String:

    • Return comp after the loop finishes.

Java Code

  1. Initialize a StringBuilder and Counter:

    • Use a StringBuilder object comp to store the compressed output.
    • Initialize a count variable to keep track of consecutive character occurrences.

  2. Loop Through the String:

    • Iterate through each character in word from the second position onward.
    • For each character, check:
      • If it’s different from the previous character.
      • If count has reached 9.
      • If it’s the last character in word.
    • If any condition is met, append the count and the previous character to comp.

  3. Update the Counter:

    • Reset count to 1 if any of the above conditions are met.
    • Otherwise, increment count for consecutive identical characters.

  4. Output the Result:

    • After the loop completes, convert comp to a string and return it.

JavaScript Code

  1. Initialize an Empty String and Counter:

    • Create an empty string comp to store the compressed result.
    • Initialize a count variable to 1 for counting character occurrences.

  2. Traverse the String:

    • Loop through each character in word starting from the second position.
    • For each character, check if:
      • It differs from the previous character.
      • count reaches 9.
      • It’s the last character in word.
    • If any condition holds true, concatenate count and the previous character to comp.

  3. Manage the Counter:

    • Reset count to 1 if the conditions are met.
    • Otherwise, increase count for each matching character.

  4. Return the Compressed Output:

    • Once the loop ends, return comp containing the compressed string.

Python Code

  1. Create a List for the Result and Initialize a Counter:

    • Use a list comp to accumulate the compressed parts, which will be joined into a string at the end.
    • Set count to 1 to count each character’s consecutive occurrences.

  2. Loop Through Each Character:

    • Iterate through word starting from the second position.
    • For each character, check:
      • If it differs from the previous character.
      • If count equals 9.
      • If it’s the last character in the string.
    • If any condition is met, add a formatted string with count and the previous character to comp.

  3. Adjust the Counter:

    • Reset count to 1 if any condition applies.
    • Otherwise, increment count for consecutive identical characters.

  4. Join and Return the Result:

    • Convert comp list to a single string using join() and return the result.

Go Code

  1. Initialize an Empty String and Counter:

    • Start with an empty string comp to hold the compressed version of word.
    • Set count to 1 to track character repetitions.

  2. Iterate Through Each Character:

    • Loop through the word string from the second character onward.
    • For each character, check:
      • If it’s different from the previous character.
      • If count equals 9.
      • If it’s the final character in the word.
    • If any condition is met, append a formatted string of count and the previous character to comp.

  3. Update the Counter:

    • Reset count to 1 if the character changes or count reaches 9.
    • Otherwise, increment count if the character repeats.

  4. Return the Final String:

    • After the loop, comp will contain the compressed result. Return comp as the output.

Summary

This approach ensures that all versions of the solution compress the string by tracking consecutive characters in groups of up to 9, adding each group's count and character to the result. By implementing the same logic across multiple languages, this guide offers a clear and consistent way to understand and solve the String Compression III problem.