🧠 Solving LeetCode Until I Become Top 1% — Day `61`

🔹 Problem: 1323. Maximum 69 Number

Difficulty: #Easy
Tags: #Greedy, #Math

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📝 Problem Summary

You’re given a positive integer made up of only digits 6 and 9. You are allowed to change at most one digit (6 → 9 or 9 → 6).
The goal is to maximize the resulting number.

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🧠 My Thought Process

• Brute Force Idea:
  Replace each 6 with 9 one by one, compute all possible numbers, and pick the maximum. This works, but it’s wasteful.

• Optimized Strategy (Greedy):

  • To maximize, we only need to flip the leftmost 6 into 9.
  • Why? Because digits on the left have higher weight in decimal representation.
  • No need to check further, just do a single replacement.

• Algorithm Used:
  
  Greedy + String Manipulation

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⚙️ Code Implementation (Python)

class Solution:
    def maximum69Number (self, num: int) -> int:
        # Convert number to string for easy manipulation
        num = str(num)

        # Replace only the first '6' with '9'
        num = num.replace('6', '9', 1)

        # Convert back to int and return
        return int(num)

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⏱️ Time & Space Complexity

• Time: O(n), where n = number of digits (string scan + replacement).
• Space: O(n) due to string conversion.

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🧩 Key Takeaways

• ✅ Greedy worked because we only need to maximize once at the highest place value.
• 💡 Learned that string manipulation can often simplify digit-based problems.
• 💭 Similar problems often involve changing the most significant digit first.

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🔁 Reflection (Self-Check)

• [x] Could I solve this without help?
• [x] Did I write code from scratch?
• [x] Did I understand why it works?
• [x] Will I be able to recall this in a week?

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🚀 Progress Tracker

Metric	Value
Day	61
Total Problems Solved	417
Confidence Today	😃
Leetcode Rating	1572