GoLang Solution with explanation
Intuition\nThe problem involves selecting k projects among given n projects. Each project has an associated profit and capital. We have to start with an initial
# Intuition\nThe problem involves selecting k projects among given n projects. Each project has an associated profit and capital. We have to start with an initial capital \'w\' and select a project whose capital is less than or equal to our current capital \'w\'. We can only select a project once. We have to maximize our profit by selecting projects in such a way that we reach the maximum profit after selecting \'k\' projects.\n\n# Approach\nTo solve this problem, we can start by creating a Project struct that will store the profit and capital of each project. We can then create a priority queue of projects sorted in descending order of their profits. This priority queue will be implemented using the heap data structure. We can push all projects with a capital less than or equal to our current capital onto this priority queue. The first project in this priority queue will have the highest profit. We can select this project, add its profit to our current capital, and repeat this process \'k\' times. If we exhaust all projects or cannot select any project because its capital is greater than our current capital, we stop and return the final value of \'w\'.\n\nWe can use the sort.Slice method to sort the projects array in increasing order of their capital. This will allow us to iterate through the projects array in a linear fashion while keeping track of the index of the last project whose capital is less than or equal to our current capital.\n\n# Complexity\n- Time complexity: $$O(nlogn)$$ because we have to sort the projects array and perform \'k\' iterations. In each iteration, we perform one heap push and one heap pop operation, both of which take $$O(logn)$$ time.\n\n- Space complexity: $$O(n)$$ because we have to store the projects array and the priority queue.\n\n# Code\n```\ntype Project struct {\n    profit int\n    capital int\n}\n\ntype PriorityQueue []Project\n\nfunc (pq PriorityQueue) Len() int {\n    return len(pq)\n}\n\nfunc (pq PriorityQueue) Less(i, j int) bool {\n    return pq[i].profit > pq[j].profit\n}\n\nfunc (pq PriorityQueue) Swap(i, j int) {\n    pq[i], pq[j] = pq[j], pq[i]\n}\n\nfunc (pq *PriorityQueue) Push(x interface{}) {\n    item := x.(Project)\n    *pq = append(*pq, item)\n}\n\nfunc (pq *PriorityQueue) Pop() interface{} {\n    old := *pq\n    n := len(old)\n    item := old[n-1]\n    *pq = old[:n-1]\n    return item\n}\n\nfunc findMaximizedCapital(k int, w int, profits []int, capital []int) int {\n    projects := make([]Project, len(profits))\n    for i := 0; i < len(profits); i++ {\n        projects[i] = Project{profit: profits[i], capital: capital[i]}\n    }\n    \n    sort.Slice(projects, func(i, j int) bool {\n        return projects[i].capital < projects[j].capital\n    })\n    \n    pq := make(PriorityQueue, 0)\n    i := 0\n    \n    for k > 0 {\n        for i < len(projects) && projects[i].capital <= w {\n            heap.Push(&pq, projects[i])\n            i++\n        }\n        \n        if len(pq) == 0 {\n            break\n        }\n        \n        p := heap.Pop(&pq).(Project)\n        w += p.profit\n        k--\n    }\n    \n    return w\n}\n```
['Heap (Priority Queue)', 'Go']