502: Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. Create a list of tuples containing (capital, profit) for each project.
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Create a list of tuples containing (capital, profit) for each project. We can use a list comprehension to achieve this.\n\n2. Sort the list of projects by increasing capital required to start them. We can use the built-in sort function with a lambda function to sort the list based on the first element (capital) of each tuple.\n\n3. Initialize a max heap to keep track of available profits from projects. We can use the built-in heapq module to create a heap.\n\n4. Loop through k projects or until we run out of projects to consider. We use a for loop with a range of k to limit the number of projects we can consider.\n\n5. Add all available projects we can currently afford to the max heap. We use a while loop to check if there are any projects left in the list and if the first project requires less capital than what we currently have. If so, we add the project to the heap.\n\n6. If there are no available projects, we can\'t make any more profit. We check if the heap is empty and if so, we break out of the loop.\n\n7. Take the project with the highest profit from the max heap and add its profit to our total. We use heapq.heappop to get the project with the highest profit (since we added negative profits to the heap), and subtract it from our current capital.\n\n8. Return the final total capital.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n        # Create a list of tuples containing (capital, profit) for each project\n        projects = [(capital[i], profits[i]) for i in range(len(profits))]\n        \n        # Sort the list of projects by increasing capital required to start them\n        projects.sort()\n        \n        # Initialize a max heap to keep track of available profits from projects\n        heap = []\n        \n        # Loop through k projects or until we run out of projects to consider\n        for _ in range(k):\n            # Add all available projects we can currently afford to the max heap\n            while projects and projects[0][0] <= w:\n                capital, profit = projects.pop(0)\n                heapq.heappush(heap, -profit)\n            \n            # If there are no available projects, we can\'t make any more profit\n            if not heap:\n                break\n            \n            # Take the project with the highest profit from the max heap and add its profit to our total\n            w -= heapq.heappop(heap)\n        \n        # Return the final total capital\n        return w\n\n```
['Array', 'Greedy', 'Sorting', 'Python', 'Python3']