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Can Be Valid Parentheses with Locked Characters

This README provides a step-by-step explanation for solving the "Can Be Valid Parentheses with Locked Characters" problem in C++, Java, JavaScript, Python, and Go. Each explanation is designed to help you understand the solution logic without directly exposing the code. Follow along to break down the approach and reasoning for each language.

🚀 C++ Code

Step-by-Step Explanation

  1. Odd Length Check:

    • First, check if the length of the string is odd. If so, return false immediately since an odd-length string can't be a valid parentheses sequence.

  2. Left-to-Right Pass:

    • Start traversing the string from left to right.
    • Maintain two variables:
      • open: Keeps track of the net count of opening brackets.
      • flexible: Counts the number of characters that can be treated as either ( or ) because they are not locked.
    • Update open based on locked characters. Increase flexible for unlocked characters.
    • If at any point open + flexible < 0, it means too many ) appear before enough (, so return false.

  3. Right-to-Left Pass:

    • Traverse the string from right to left.
    • Reuse open to count the net number of closing brackets ()).
    • Similarly, use flexible for unlocked characters.
    • If open + flexible < 0, return false.

  4. Return Result:

    • If both passes succeed, return true.

🚀 Java Code

Step-by-Step Explanation

  1. Odd Length Check:

    • Verify if the length of the string is odd. Return false if it is because a valid parentheses sequence must have an even number of characters.

  2. Left-to-Right Pass:

    • Traverse the string from left to right.
    • Maintain:
      • open: Keeps track of the count of ( brackets.
      • flexible: Tracks the number of unlocked characters.
    • Update open for locked characters (locked = 1), and increase flexible for unlocked ones.
    • If open + flexible < 0, return false as too many ) have appeared.

  3. Right-to-Left Pass:

    • Perform a reverse traversal of the string.
    • Use open to count closing brackets ()), and flexible for unlocked characters.
    • If open + flexible < 0, return false.

  4. Return Final Result:

    • If both passes succeed, return true.

🚀 JavaScript Code

Step-by-Step Explanation

  1. Odd Length Check:

    • Check if the string's length is odd. If yes, return false since balancing is impossible.

  2. Left-to-Right Pass:

    • Traverse the string from the beginning to the end.
    • Maintain:
      • open: Tracks the balance of opening brackets.
      • flexible: Tracks characters that are not locked.
    • Update the counts based on whether characters are locked or unlocked.
    • If open + flexible < 0, return false.

  3. Right-to-Left Pass:

    • Reverse the traversal of the string.
    • Reuse open for closing brackets and flexible for unlocked characters.
    • If open + flexible < 0, return false.

  4. Return Result:

    • If both checks pass, return true.

🚀 Python Code

Step-by-Step Explanation

  1. Odd Length Check:

    • Start by checking if the string has an odd number of characters. If true, immediately return False.

  2. Left-to-Right Pass:

    • Traverse the string from left to right.
    • Maintain two counters:
      • open: Tracks the net count of opening brackets.
      • flexible: Counts characters that are unlocked (locked = '0').
    • Increment open for ( and decrement it for ) (locked characters). Increment flexible for unlocked characters.
    • If at any point open + flexible < 0, return False.

  3. Right-to-Left Pass:

    • Traverse the string in reverse.
    • Recalculate open for closing brackets ()).
    • Use flexible to handle unlocked characters.
    • If open + flexible < 0, return False.

  4. Return the Result:

    • If both passes validate the string, return True.

🚀 Go Code

Step-by-Step Explanation

  1. Odd Length Check:

    • Check if the length of the string is odd. If it is, return false since balancing is impossible.

  2. Left-to-Right Pass:

    • Start from the beginning and traverse to the end.
    • Keep two variables:
      • open: Tracks the net count of opening brackets.
      • flexible: Tracks the number of characters that can be either ( or ) based on their unlocked state.
    • Update open for locked characters and flexible for unlocked ones.
    • If open + flexible < 0 at any point, return false.

  3. Right-to-Left Pass:

    • Traverse the string from the end to the beginning.
    • Use open for closing brackets and flexible for unlocked characters.
    • If open + flexible < 0, return false.

  4. Final Result:

    • If both passes confirm the string is valid, return true.

🌟 Common Logic Across All Languages

  • Perform two traversals (left-to-right and right-to-left) to validate the balance of parentheses while accommodating unlocked characters.
  • Use simple counters (open and flexible) to dynamically manage the balance as you iterate.
  • Return true only if both traversals succeed without imbalance.

✨ Key Takeaways

  • This problem demonstrates the importance of two-pass traversal and dynamic updates based on constraints (locked vs. unlocked).
  • While the syntax and structure differ slightly across languages, the core logic remains the same.

Enjoy solving! 🚀