## Intuition The problem requires us to shorten each word in a list such that the shortened versions are unique and have a length between 1 and 4 characters. The challenge is to ensure that no two shortened words are the same, which involves finding a unique combination of characters for each word. ## Approach 1. **Initial Check**: If any word is already 4 characters or less, we can use it as is. Otherwise, we need to generate a shortened version. 2. **Generate Candidates**: For each word longer than 4 characters, generate all possible substrings of lengths 1 to 4. These substrings are potential shortened versions. 3. **Unique Assignment**: Use a greedy approach to assign a unique shortened version to each word: - Start with the shortest possible substrings (length 1) and move up to length 4. - Use a set to track already used shortened versions to ensure uniqueness. - If a word cannot be assigned a unique shortened version, output `-1`. 4. **Output**: If all words can be assigned a unique shortened version, print them. Otherwise, print `-1`. ## Complexity - **Time**: O(n * 2^10), where n is the number of words. This is because generating all possible substrings of lengths 1 to 4 for each word is manageable given the constraints. - **Space**: O(n), primarily for storing the shortened versions and tracking used substrings. ## Solution (C++) ```cpp #include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n; cin >> n; vector words(n); for (int i = 0; i < n; ++i) { cin >> words[i]; } unordered_set used; vector result(n); for (int i = 0; i < n; ++i) { string word = words[i]; int len = word.size(); bool found = false; for (int l = 1; l <= 4 && !found; ++l) { for (int j = 0; j <= len - l; ++j) { string candidate = word.substr(j, l); if (used.find(candidate) == used.end()) { result[i] = candidate; used.insert(candidate); found = true; break; } } } if (!found) { cout << -1 << endl; return 0; } } for (const string &shortened : result) { cout << shortened << endl; } return 0; } ``` This solution efficiently assigns a unique shortened version to each word by iterating over possible substrings and checking for uniqueness using a set. If a unique assignment is not possible, it outputs `-1`.

Brevity is Soul of Wit

Problem Details