Daily Challenge is Good
Solution Approach\nCombine and Sort Projects:\n\nCreate a list of projects, where each project is represented by its required capital and profit.\nSort this lis
Solution Approach\nCombine and Sort Projects:\n\nCreate a list of projects, where each project is represented by its required capital and profit.\nSort this list based on the capital required to start the projects.\nUse a Max-Heap for Profits:\n\nUse a max-heap to keep track of the profits of the projects that can be started with the current capital.\nThis helps us quickly select the most profitable project we can afford.\nSelect Projects Iteratively:\n\nFor up to k iterations, add all projects that can be started with the current capital to the max-heap.\nIf no projects can be started (the heap is empty), break the loop.\nOtherwise, pick the project with the highest profit from the heap and increase the capital by this profit.\nReturn the Final Capital:\n\nAfter completing up to k projects, return the final capital.\n\nJava\n```\npublic class Solution {\n    public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n        int n = profits.length;\n    int[][] projects =new int[n][2];\n        for (int i = 0; i < n; i++) {\n            projects[i][0] =capital[i];\n            projects[i][1] =profits[i];\n\n        }        \n        Arrays.sort(projects, (a, b) -> Integer.compare(a[0], b[0]));\n\n        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) ->Integer.compare(b, a));        \n        int currentCapital = w;\n        int projectIndex = 0;\n        for (int i = 0; i < k; i++) {\n\n            while (projectIndex < n && projects[projectIndex][0] <= currentCapital) {\n                maxHeap.add(projects[projectIndex][1]);\n                projectIndex++;\n        }\n            if (maxHeap.isEmpty()) {\n                break;\n            }\n            currentCapital += maxHeap.poll();\n\n        }\n        return currentCapital;\n    }\n}\n\n\n```\n\nC\n```\ntypedef struct {\n    int capital;\n    int profit;\n} Project;\n\nint compareProjects(const void* a, const void* b) {\n    return ((Project*)a)->capital - ((Project*)b)->capital;\n}\n\ntypedef struct {\n    int* data;\n    int size;\n    int capacity;\n} MaxHeap;\n\nMaxHeap* createMaxHeap(int capacity) {\n    MaxHeap* heap = (MaxHeap*)malloc(sizeof(MaxHeap));\n    heap->data = (int*)malloc(capacity * sizeof(int));\n    heap->size = 0;\n    heap->capacity = capacity;\n    return heap;\n}\n\nvoid swap(int* a, int* b) {\n    int temp = *a;\n    *a = *b;\n    *b = temp;\n}\n\nvoid maxHeapifyUp(MaxHeap* heap, int index) {\n    while (index > 0 && heap->data[index] > heap->data[(index - 1) / 2]) {\n        swap(&heap->data[index], &heap->data[(index - 1) / 2]);\n        index = (index - 1) / 2;\n    }\n}\n\nvoid maxHeapifyDown(MaxHeap* heap, int index) {\n    int largest = index;\n    int left = 2 * index + 1;\n    int right = 2 * index + 2;\n    \n    if (left < heap->size && heap->data[left] > heap->data[largest]) {\n        largest = left;\n    }\n    if (right < heap->size && heap->data[right] > heap->data[largest]) {\n        largest = right;\n    }\n    if (largest != index) {\n        swap(&heap->data[index], &heap->data[largest]);\n        maxHeapifyDown(heap, largest);\n    }\n}\n\nvoid insertMaxHeap(MaxHeap* heap, int value) {\n    heap->data[heap->size] = value;\n    heap->size++;\n    maxHeapifyUp(heap, heap->size - 1);\n}\n\nint extractMax(MaxHeap* heap) {\n    if (heap->size == 0) {\n        return 0;\n    }\n    int maxValue = heap->data[0];\n    heap->data[0] = heap->data[heap->size - 1];\n    heap->size--;\n    maxHeapifyDown(heap, 0);\n    return maxValue;\n}\n\nint findMaximizedCapital(int k, int w, int* profits, int* capital, int n) {\n    Project* projects = (Project*)malloc(n * sizeof(Project));\n    for (int i = 0; i < n; i++) {\n        projects[i].capital = capital[i];\n        projects[i].profit = profits[i];\n    }\n    \n    qsort(projects, n, sizeof(Project), compareProjects);\n    \n    MaxHeap* maxHeap = createMaxHeap(n);\n    int currentCapital = w;\n    int projectIndex = 0;\n    \n    for (int i = 0; i < k; i++) {\n        while (projectIndex < n && projects[projectIndex].capital <= currentCapital) {\n            insertMaxHeap(maxHeap, projects[projectIndex].profit);\n            projectIndex++;\n        }\n        \n        if (maxHeap->size == 0) {\n            break;\n        }\n        \n        currentCapital += extractMax(maxHeap);\n    }\n    \n    free(projects);\n    free(maxHeap->data);\n    free(maxHeap);\n    return currentCapital;\n}\n\n```\n\nPython\n```\nimport heapq\n\nclass Solution:\n    def findMaximizedCapital(self, k: int, w: int, profits: list[int], capital: list[int]) -> int:\n        n = len(profits)\n        projects = [(capital[i], profits[i]) for i in range(n)]\n        projects.sort()\n        \n        max_heap = []\n        current_capital = w\n        project_index = 0\n        \n        for _ in range(k):\n            while project_index < n and projects[project_index][0] <= current_capital:\n                heapq.heappush(max_heap, -projects[project_index][1])\n                project_index += 1\n            \n            if not max_heap:\n                break\n            \n            current_capital += -heapq.heappop(max_heap)\n        \n        return current_capital\n\n```
['Greedy', 'C', 'Heap (Priority Queue)', 'Python', 'Java']