📘 Day 15 – Teacher's Day Challenge | Boy or Girl

🗓️ Date: August 08, 2025
🧩 Platform: Codeforces
🔢 Problems Solved Today: 1
🎯 Focus Topic: String Manipulation, Sets

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🔗 Table of Contents

• Problem 1 – Boy or Girl

• Daily Summary

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🧩 Problem 1 – Boy or Girl

📚 Difficulty: Easy-Medium
🧠 Concepts: String Manipulation, Unique Characters

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📄 Problem Statement (Summary):
Given a username string consisting of lowercase English letters, determine the "gender" based on the count of distinct characters:

• If the count is odd → Male → print "IGNORE HIM!"

• If the count is even → Female → print "CHAT WITH HER!"

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🔢 Input Format:
A single string containing only lowercase English letters (length ≤ 100).

🧾 Output Format:
One of the following strings:

• "CHAT WITH HER!" (female)

• "IGNORE HIM!" (male)

🔍 Example 1:
Input:
wjmzbmr
Output:
CHAT WITH HER!

🔍 Example 2:
Input:
xiaodao
Output:
IGNORE HIM!

🔍 Example 3:
Input:
sevenkplus
Output:
CHAT WITH HER!

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💡 Approach:

• Sort the string to group identical letters.

• Remove duplicates to find the count of distinct characters.

• Check if count is odd or even.

• Output result accordingly.

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🧪 Code (C++):

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    string str;
    cin >> str;
    int i = 0, j;
    sort(str.begin(), str.end());
    for (j = 1; j < str.length(); j++)
    {
        if (str[i] == str[j])
        {
            str.erase(j, 1);
            j--;
            i--;
        }
        i++;
    }
    if (str.length() & 1)
        cout << "IGNORE HIM!";
    else
        cout << "CHAT WITH HER!";
    return 0;
}

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📸 Submission Screenshot:

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✅ Key Takeaways:

• Sorting helps in grouping duplicates for easy removal.

• Distinct character counting is a common string problem.

• Simple logic with % 2 can determine parity-based decisions.

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📈 Daily Summary

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🏷️ Tags:
#codeforces #strings #cpp #beginner #dsa #43DaysChallenge