Python3, Fully Explained with comments|| Beats 97%🔥|| Sorting and Two-Pointer Technique
Intuition\n Describe your first thoughts on how to solve this problem. \n## Sorting and Two-Pointer Technique:\n1. Sort the worker array to process workers in i
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n## **Sorting and Two-Pointer Technique:**\n1. Sort the `worker` array to process workers in increasing order of their abilities.    \n2. Sort the `difficulty` and `profit` arrays together based on difficulty.     \n3. Use a two-pointer technique to track the maximum profit a worker can achieve by iterating through the sorted tasks and keeping track of the highest profit seen so far.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n### 1. Initialization:\n\n-  `max_profit` is initialized to 0 to keep track of the total profit\n- The `worker` list is sorted to process workers in order of their abilities.\n- The `difficulty` and `profit` arrays are zipped together and sorted based on difficulty.\n\n### 2. Two-Pointer Technique:\n\n- A pointer `i` is used to iterate through the sorted     `difficulty-profit` pairs.\n- `cur_profit` is used to keep track of the maximum profit seen so far.\n\n### 3. Iterate Through Workers:\n\n- For each worker\'s ability:\n    - The inner while loop checks if the current worker can handle the task at `vals[i]`.\n    - If the worker can handle the task, update `cur_profit` to the maximum profit of the tasks they can perform.\n    - Increment `i` to move to the next task.\n- After processing all tasks the current worker can handle, add `cur_profit` to `max_profit`.\n### 4. Return Result:\n\n- After processing all workers, return the total `max_profit`.\n\n# Complexity\n### - Time complexity:\n  - Sorting the worker array: ******O(WlogW)******\n  - Sorting the combined difficulty and profit arrays: ****O(DlogD)****\n  - Iterating through the workers and tasks: ****O(W+D)****\n  - *Overall*: ****O((W+D)logD+W+D)****\n\n### - Space complexity:\n- ****O(D)**** for storing the combined list of difficulty and profit.\n# Conclusion :\n- This approach ensures that we efficiently maximize the profit by leveraging sorting and a two-pointer technique, making it both time and space efficient for large inputs.\n\n# Code\n```\nclass Solution:\n    def maxProfitAssignment(self, difficulty: List[int], profit: List[int], worker: List[int]) -> int:\n        max_profit = 0  # Initialize the total maximum profit to 0\n        worker.sort()  # Sort the worker abilities in ascending order\n        vals = sorted(zip(difficulty, profit))  # Combine and sort difficulty and profit by difficulty\n        \n        i = 0  # Pointer for iterating through the sorted difficulty-profit pairs\n        cur_profit = 0  # Variable to track the maximum profit achievable at the current point\n        \n        # Iterate through each worker\'s ability\n        for ability in worker:\n            # Update cur_profit with the highest profit for tasks the worker can perform\n            while i < len(difficulty) and ability >= vals[i][0]:\n                cur_profit = max(cur_profit, vals[i][1])  # Update cur_profit to the highest profit seen so far\n                i += 1  # Move to the next task\n            \n            max_profit += cur_profit  # Add the best possible profit for the current worker to max_profit\n        \n        return max_profit  # Return the total maximum profit\n\n\n```
['Array', 'Two Pointers', 'Sorting', 'Python3']