🧠 Solving LeetCode Until I Become Top 1% — Day `74`

🔹 Problem: 3516 Find Closest Person

Difficulty: #Easy
Tags: #Math, #Comparison

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📝 Problem Summary

First of all, maybe the EASIEST PROBLEM IN LEETCODE HISTORY.

You are given three integers x, y, and z. The task is to determine which of x or y is closer to z.

• If both are equally close → return 0.
• If x is closer → return 1.
• If y is closer → return 2.

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🧠 My Thought Process

• Brute Force Idea:
  I think you can solve this eeeasily with simple math.

• Optimized Strategy:
  The brute force is optimal here — no loops, no data structures, just math.

  • Compute |x - z| and |y - z|.
  • Compare the two distances.
  • Decide the answer based on which difference is smaller.

• Algorithm Used:
  
  Pure mathematical comparison using absolute difference.

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⚙️ Code Implementation (Python)

class Solution:
    def findClosest(self, x: int, y: int, z: int) -> int:
        x_abs = abs(x - z)
        y_abs = abs(y - z)

        if x_abs == y_abs:
            return 0
        elif x_abs < y_abs:
            return 1
        else:
            return 2

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⏱️ Time & Space Complexity

• Time: O(1) → Only a few arithmetic operations.
• Space: O(1) → No extra memory used.

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🧩 Key Takeaways

• ✅ Learned how simple absolute difference comparisons solve closest-number problems.
• 💡 The trick is recognizing that absolute difference directly measures closeness.
• 💭 This technique can be reused in problems like “find closest element to a target in an array” (but with loops/binary search).

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🔁 Reflection (Self-Check)

• [x] Could I solve this without help? (Yes, straightforward math)
• [x] Did I write code from scratch? (Yes)
• [x] Did I understand why it works? (Yes, absolute difference logic)
• [x] Will I be able to recall this in a week? (Yes, very simple pattern)

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🚀 Progress Tracker

Metric	Value
Day	74
Total Problems Solved	437
Confidence Today	😃
Leetcode Rating	1530