# Problem Understanding We are given a list of agents, each with a strength rating. A task force is considered valid when it includes a group of agents (with at least one member) such that, if the smallest strength in the group is $s$ and there are $k$ agents in the group, then the product $$s \times k$$ is at least $x$. An agent may only be used once and we do not need to use every agent. The goal is to form as many valid task forces as possible. The key observation is that the smallest strength in any chosen group determines whether the product requirement is met. Since the product condition is $s \times k \ge x$, it implies that if an agent has strength $s$, then you need at least $k \ge \lceil \frac{x}{s} \rceil$ agents in the group that includes that weakest agent. # Solution Analysis A greedy algorithm works well for this problem: 1. **Sort the Agents by Strength:** It is ideal to sort the list of agents in descending order of strength. This is because a stronger agent has a lower requirement on the number of agents needed to meet the product $s \times k$. Processing from greatest to smallest simplifies grouping; while grouping, you are always able to meet the threshold condition sooner if you use stronger agents. 2. **Greedy Group Formation:** Iterate through the sorted list and try to form a group: - Maintain a counter (let’s call it `groupSize`) for the number of agents in the current potential group. - For each agent in the sorted list, increment the counter. - Check if $s \times groupSize \ge x$, where $s$ is the current agent’s strength (which is the smallest strength in the group because the list is in descending order). - If the condition is met, then you have formed a valid task force. Increase your answer count for task forces and reset the `groupSize` counter to start forming a new group. - Continue the process until all agents have been considered. 3. **Rationale Behind the Greedy Choice:** Consider the strongest agents first; these agents require fewer additional agents to meet the threshold. If you try to form groups starting with weaker agents, you might end up needing many more agents, which could be better allocated to satisfy the condition in other groups that already have strong candidates. # Time and Space Complexity Analysis - **Time Complexity:** - Sorting the array takes $O(n \log n)$ for each test case. - Processing each agent in a single pass contributes an additional $O(n)$. - Since the sum of all $n$ across test cases is at most $10^5$, the overall complexity remains efficient. - **Space Complexity:** - The algorithm uses $O(n)$ space to store the list of agents. - Additional space usage is constant ($O(1)$) for counters and variables. This ensures that the algorithm efficiently handles the input constraints.