Submission

• string least diff = delete + insertion
• 1st case: least diff = k
• 2nd case: least diff < k, but k – least_dff is divisible by 2, we can exhaustively append one letter and then remove one after achieving the final state of 1st case
• 1st + 2nd case: least_diff <= k && (k – least_diff) is divisible by 2
• 3rd case: k is no smaller than the sum of the length of the 2 strings, we can remove a whole string and construct an identical string to the other. When the string is empty, we can use up the extra steps by trying to remove a letter from the empty string.

#include <bits/stdc++.h>

using namespace std;

// Complete the appendAndDelete function below.
string appendAndDelete(string s, string t, int k) {
    int i;
    bool isDiff = true;
    for(i = 0; i != s.size() && i != t.size() ; i++) {
        if(s[i] != t[i]) {
            break;
        }
    }
    int d = s.size() + t.size() - 2 * i;
    if(k >= d && (d - k) % 2 == 0) return "Yes";
    if(k > s.size() + t.size()) return "Yes";
    return "No";
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    string s;
    getline(cin, s);

    string t;
    getline(cin, t);

    int k;
    cin >> k;
    cin.ignore(numeric_limits<streamsize>::max(), '\n');

    string result = appendAndDelete(s, t, k);

    fout << result << "\n";

    fout.close();

    return 0;
}