C# Solution for IPO Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this solution is to efficiently manage the selection of projects u
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to efficiently manage the selection of projects using two heaps (priority queues) to always have access to the most profitable project that can be started given the current available capital. By separating the projects based on whether they can be started or not, we ensure that we always pick the best possible project at each step.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tInitialize Priority Queues:\n\t\u2022\tMin-Heap for Capital Requirements: This heap helps in quickly identifying projects that can be unlocked based on the current available capital.\n\t\u2022\tMax-Heap for Profits: This heap helps in selecting the most profitable project that can currently be started.\n2.\tInsert Projects into Min-Heap:\n\t\u2022\tInsert all projects into the min-heap, sorted by their capital requirements. This allows us to efficiently determine which projects become available as our capital increases.\n3.\tIterate for k Projects:\n\t\u2022\tFor each of the k iterations:\n\t\u2022\tTransfer all projects from the min-heap to the max-heap that can be started with the current capital.\n\t\u2022\tIf no projects can be started (i.e., the max-heap is empty), break the loop.\n\t\u2022\tSelect the project with the highest profit from the max-heap and update the current capital by adding the profit of the selected project.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tInserting Projects into Min-Heap: Each project insertion into the min-heap takes O(log n). Since there are n projects, this step takes O(nlog n).\n\u2022\tIterating and Managing Heaps: For each of the k iterations:\n\u2022\tMoving projects from the min-heap to the max-heap involves heap operations that are O(log n) each.\n\u2022\tIn the worst case, all n projects are moved once, leading to O(nlog n).\n\u2022\tRemoving the most profitable project from the max-heap is O(log n).\nSince these operations are performed up to k times, this part of the process has a complexity of O(klog n).\n\u2022\tTotal Time Complexity: The overall time complexity is O(nlog n + klog n). Given that k can be at most n, the time complexity can be simplified to O(nlog n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tMin-Heap and Max-Heap: Both heaps together store all projects, requiring O(n) space.\n\u2022\tAuxiliary Space: Additional space is used for storing variables and temporary data, but it is dominated by the space required for the heaps.\n\u2022\tTotal Space Complexity: O(n)\n\n# Code\n```\npublic class Solution {\n    public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n        int n = profits.Length;\n        \n        // Min-heap for capital requirements (sorted by capital)\n        PriorityQueue<(int capital, int profit), int> minHeap = new PriorityQueue<(int, int), int>();\n        // Max-heap for profits (sorted by profit)\n        PriorityQueue<int, int> maxHeap = new PriorityQueue<int, int>(Comparer<int>.Create((a, b) => b.CompareTo(a)));\n        \n        // Insert all projects into the min-heap\n        for (int i = 0; i < n; i++) {\n            minHeap.Enqueue((capital[i], profits[i]), capital[i]);\n        }\n        \n        int currentCapital = w;\n        \n        for (int i = 0; i < k; i++) {\n            // Move projects that can be unlocked with the current capital to the max-heap\n            while (minHeap.Count > 0 && minHeap.Peek().capital <= currentCapital) {\n                var project = minHeap.Dequeue();\n                maxHeap.Enqueue(project.profit, project.profit);\n            }\n            \n            // If no projects can be started, break\n            if (maxHeap.Count == 0) {\n                break;\n            }\n            \n            // Select the project with the maximum profit\n            currentCapital += maxHeap.Dequeue();\n        }\n        \n        return currentCapital;\n    }\n}\n```
['C#']