O(NLOGN) || In depth explanation || Step Wise
\n\nExplanation with Visualizations:\n\nProblem:\n\nThe provided code aims to solve the maximum profit assignment problem. Given arrays representing job difficu
\n\n**Explanation with Visualizations:**\n\n**Problem:**\n\nThe provided code aims to solve the maximum profit assignment problem. Given arrays representing job difficulties (`difficulty`), profits (`profit`), and worker skills (`worker`), the goal is to assign jobs to workers in a way that maximizes the total accumulated profit. A worker can only complete a job if their skill level is greater than or equal to the job\'s difficulty.\n\n**Approach:**\n\nThe code leverages a priority queue and sorting to efficiently determine the optimal job assignments:\n\n1. **PriorityQueue for Jobs:**\n   - `PriorityQueue<Pair> pq = new PriorityQueue<>((a, b) -> b.profit - a.profit);`:\n     - Creates a priority queue `pq` to store job information as `Pair` objects (profit, difficulty).\n     - The custom comparator ensures descending order by profit. This prioritizes jobs with potentially higher rewards during assignment.\n   - `pq.add(new Pair(0, 0));`: Adds a dummy entry (zero profit, zero difficulty) to handle cases where workers might not be assigned any job (empty queue).\n\n2. **Adding Jobs to Queue:**\n   - `for (int i = 0; i < profit.length; i++) { ... }`: Iterates through the `profit` and `difficulty` arrays.\n     - `pq.add(new Pair(profit[i], difficulty[i]));`: Creates a `Pair` for each job and adds it to the `pq`. This builds a queue of jobs sorted by descending profit.\n\n3. **Sorting Workers:**\n   - `Arrays.sort(worker);`: Sorts the `worker` array in ascending order. This ensures we consider workers with lower skill levels first during the assignment process.\n\n4. **Assigning Jobs to Workers:**\n   - `for (int i = worker.length - 1; i >= 0; i--) { ... }`: Iterates through the `worker` array in reverse order (starting from the most skilled worker).\n     - `while (pq.peek().difficulty > worker[i]) { ... }`: This loop keeps removing jobs from `pq` using `poll()` as long as their difficulty exceeds the current worker\'s skill level. This guarantees job suitability based on skill constraints.\n     - `ans += pq.peek().profit;`: Once a suitable job is found (peek at the queue), its profit is added to the `ans` variable (total accumulated profit).\n\n**Visualizations (incorporating strengths from both responses):**\n\n* **Job Queue (Priority Queue):** Imagine a prioritized list where jobs with higher profits are at the front. As jobs are considered for assignment, the queue is modified dynamically.\n\n```\n   +-----------+-----------+\n   | Profit: 10 | Difficulty: 5 | (Highest priority)\n   +-----------+-----------+\n   | Profit: 8  | Difficulty: 3 |\n   +-----------+-----------+\n   | Profit: 5  | Difficulty: 2 |\n   +-----------+-----------+\n   | ...        | ...           |\n   +-----------+-----------+\n   | Profit: 2  | Difficulty: 1 | (Lowest priority)\n   +-----------+-----------+\n   | Profit: 0  | Difficulty: 0 | (Dummy entry)\n   +-----------+-----------+\n```\n\n* **Worker Skill Levels:** Picture a sorted list where workers with lower skills are at the beginning.\n\n```\n[Worker 1 skill, Worker 2 skill, ..., Worker N skill] (Sorted in ascending order)\n```\n\n**Assignment Process:**\n\n- The code iterates through the workers in descending skill order.\n- For each worker, it checks the `pq` (job queue).\n  - If the job\'s difficulty is higher than the worker\'s skill, the job is removed from `pq` using `poll()`. This ensures only suitable jobs are considered.\n- Once a suitable job is found (top of `pq`), its profit is added to the accumulated profit (`ans`).\n- The process continues until all workers have been considered.\n\n**Key Points:**\n\n- Sorting workers and using a descending priority queue for jobs lead to efficient assignment.\n- The dummy entry in `pq` handles cases where no job is assignable to a worker.\n- The code prioritizes assigning jobs with higher potential rewards to maximize total profit.\n\n**Complexity Analysis:**\n\n- Time Complexity:\n  - O(n log n) for adding jobs to the priority queue (heap operations).\n  - O(m log m) for sorting the worker skills.\n  - O(m * log k) for iterating through workers and potentially\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(NlogN)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(N)\n# Code\n```\nclass Pair{\n    int profit;\n    int difficulty;\n    Pair(int profit, int difficulty) {\n    this.profit = profit;\n    this.difficulty = difficulty;\n    }\n  \n}\n\nclass Solution {\n    public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {\n        PriorityQueue<Pair> pq=new PriorityQueue<>((a,b)-> b.profit-a.profit);\n        pq.add(new Pair(0,0));\n        for(int i=0;i<profit.length;i++){\n            pq.add(new Pair(profit[i],difficulty[i]));\n        }\n        int ans=0;\n        Arrays.sort(worker);\n        for(int i=worker.length-1;i>=0;i--){\n            while(pq.peek().difficulty>worker[i]){\n                pq.poll();\n            }\n            ans+=pq.peek().profit;\n        }\n        return ans;\n    }\n}\n```
['Sorting', 'Heap (Priority Queue)', 'Java']