📘 Day 1 – Teacher's Day Challenge | Even Digit Count + Two Sum

🗓️ Date: July 25, 2025
🔢 Problems Solved Today: 2
🎯 Focus Topic: Arrays, Brute Force, Digit Operations

💻 Platform: Leetcode

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🔗 Table of Contents

• Problem 1 – Count Numbers with Even Number of Digits

• Problem 2 – Two Sum

• Daily Summary

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🧩 Problem 1 – Count Numbers with Even Number of Digits

📚 Difficulty: Easy
🧠 Concepts: Math, Digit Count

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📄 Problem Statement (Summary):

Given an array nums, return how many numbers have an even number of digits.

Example 1:
Input: [12,345,2,6,7896] → Output: 2

Example 2:
Input: [555,901,482,1771] → Output: 1

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💡 Approach:

• Count the digits in each number using logarithm

• If the count is even, increment result counter

• Edge case: 0 has 1 digit

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🧪 Code (C++):

class Solution {
public:
    int countDigits(int n) {
        if (n == 0)
            return 1; // Handle 0 separately
        return (int)(log10(abs(n))) + 1;
    }

    int findNumbers(vector<int>& nums) {
        int c = 0;
        for (int num : nums)
            if (!(countDigits(num) & 1))  // Check if digit count is even
                c++;
        return c;
    }
};

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📸 Submission Screenshot:

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✅ Key Takeaways:

• log10(n) + 1 gives digit count for positive numbers

• Bitwise & 1 helps quickly check for odd/even

• Edge case handled for 0 separately

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🧩 Problem 2 – Two Sum

📚 Difficulty: Easy
🧠 Concepts: Brute Force, Array Traversal

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📄 Problem Statement (Summary):

Find two indices in an array such that their values add up to a given target.
Return the indices in any order.

Example:
Input: nums = [2,7,11,15], target = 9 → Output: [0,1]

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💡 Approach:

• Brute force using nested loops to try every pair

• Return the first valid pair that sums to the target

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🧪 Code (C++):

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int sum = 0;
        vector<int> sum_arr;
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                sum = nums[j] + nums[i];
                if (sum == target) {
                    sum_arr.push_back(i);
                    sum_arr.push_back(j);
                    return sum_arr;
                }
            }
        }
        return sum_arr;
    }
};

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📸 Submission Screenshot:

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✅ Key Takeaways:

• Brute force is simple but O(n²) in time

• Better version uses HashMap (O(n)), will explore later

• Ensure not using same element twice

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📈 Daily Summary

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🏷️ Tags:

#leetcode #43DaysChallenge #arrays #c++ #dsa #twosum #evenDigits