Detailed Solution With Steps
\nclass Solution {\npublic static int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n    int n = profits.length;\n\n    // Create a list of
```\nclass Solution {\npublic static int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n    int n = profits.length;\n\n    // Create a list of pairs (capital, profit) for all n projects\n    List<int[]> projects = new ArrayList<>();\n    for (int i = 0; i < n; i++) {\n        projects.add(new int[]{capital[i], profits[i]});\n    }\n\n    // Sort the list of projects by their minimum capital requirements in ascending order\n    projects.sort((a, b) -> a[0] - b[0]);\n\n    // Initialize a priority queue to store the profits of the available projects\n    PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());\n\n    // Initialize a variable i to 0 to keep track of the current project\n    int i = 0;\n\n    // Iterate over k projects\n    while (k > 0) {\n        // While i is less than n and the minimum capital requirement of the current project is less than or equal to w\n        while (i < n && projects.get(i)[0] <= w) {\n            // Add the profit of the current project to the priority queue\n            pq.offer(projects.get(i)[1]);\n            i++;\n        }\n\n        // If the priority queue is empty, break out of the loop because we can\'t afford any more projects\n        if (pq.isEmpty()) {\n            break;\n        }\n\n        // Select the project with the highest profit from the priority queue and add its profit to w\n        w += pq.poll();\n        k--;\n    }\n\n    return w;\n}\n}\n```\n# Upvoting is much appreciated
['Java']