Merge pull request #1963 from katarianikita2003/main

Create 0502-IPO.md

dsa-solutions/lc-solutions/0500-0599/0502-IPO.md

@@ -0,0 +1,292 @@
+---
+id: IPO
+title: IPO
+sidebar_label: IPO
+tags: 
+    - Dynamic Programming
+    - Greedy Algorithm
+    - Priority Queue
+---
+
+## Problem Description
+
+| Problem Statement                                       | Solution Link                                                              | LeetCode Profile                                        |
+| :------------------------------------------------------ | :------------------------------------------------------------------------- | :------------------------------------------------------ |
+| [IPO](https://leetcode.com/problems/IPO/description/) | [IPO Solution on LeetCode](https://leetcode.com/problems/IPO/solutions/) | [Nikita Saini](https://leetcode.com/u/Saini_Nikita/) |
+
+## Problem Description
+
+You are given `n` projects where the ith project has a pure profit `profits[i]` and a minimum capital of `capital[i]` is needed to start it. Initially, you have `w` capital. When you finish a project, you obtain its pure profit, which is added to your total capital.
+
+Your goal is to pick a list of at most `k` distinct projects to maximize your final capital.
+
+## Constraints
+
+- `1 <= k <= 10^5`
+- `0 <= w <= 10^9`
+- `1 <= n <= 10^5`
+- `0 <= profits[i] <= 10^4`
+- `0 <= capital[i] <= 10^9`
+
+## Approach
+
+The problem can be approached using a combination of greedy strategy and efficient data structures:
+1. **Priority Queue (Max-Heap)**: Use a max-heap to always select the project with the maximum profit that can be started with the current capital.
+2. **Sorting**: Sort projects based on their capital requirements.
+3. **Iterative Selection**: Iterate up to `k` times (or until no more projects can be started) to select the project with the highest profit that can be started with the current capital.
+
+### Solution in Different languages
+
+## Solution in Python
+
+```python
+import heapq
+
+def findMaximizedCapital(k, w, profits, capital):
+    n = len(profits)
+    projects = sorted(zip(capital, profits))  # Sort projects by their capital requirements
+    available_projects = []
+    idx = 0
+    for _ in range(k):
+        while idx < n and projects[idx][0] <= w:
+            heapq.heappush(available_projects, -projects[idx][1])  # Max-heap for profits
+            idx += 1
+        if available_projects:
+            w -= heapq.heappop(available_projects)  # Add the profit from the best project
+        else:
+            break
+    return w
+
+# Example usage:
+k = 2
+w = 0
+profits = [1, 2, 3]
+capital = [0, 1, 1]
+print(findMaximizedCapital(k, w, profits, capital))  # Output: 4
+```
+
+## Solution in Java
+
+```java
+import java.util.PriorityQueue;
+
+class Solution {
+    public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {
+        int n = profits.length;
+        int[][] projects = new int[n][2];
+        for (int i = 0; i < n; i++) {
+            projects[i][0] = capital[i];
+            projects[i][1] = profits[i];
+        }
+        Arrays.sort(projects, (a, b) -> a[0] - b[0]); // Sort projects by their capital requirements
+        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
+        int idx = 0;
+        for (int i = 0; i < k; i++) {
+            while (idx < n && projects[idx][0] <= w) {
+                maxHeap.offer(projects[idx][1]); // Max-heap for profits
+                idx++;
+            }
+            if (!maxHeap.isEmpty()) {
+                w += maxHeap.poll(); // Add the profit from the best project
+            } else {
+                break;
+            }
+        }
+        return w;
+    }
+}
+```
+
+## Solution in C++
+
+```cpp
+#include <vector>
+#include <queue>
+#include <algorithm>
+using namespace std;
+
+class Solution {
+public:
+    int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {
+        int n = profits.size();
+        vector<pair<int, int>> projects(n);
+        for (int i = 0; i < n; ++i) {
+            projects[i] = {capital[i], profits[i]};
+        }
+        sort(projects.begin(), projects.end()); // Sort projects by their capital requirements
+        priority_queue<int> maxHeap;
+        int idx = 0;
+        for (int i = 0; i < k; ++i) {
+            while (idx < n && projects[idx].first <= w) {
+                maxHeap.push(projects[idx].second); // Max-heap for profits
+                ++idx;
+            }
+            if (!maxHeap.empty()) {
+                w += maxHeap.top(); // Add the profit from the best project
+                maxHeap.pop();
+            } else {
+                break;
+            }
+        }
+        return w;
+    }
+};
+```
+
+## Solution in C
+
+```c
+#include <stdlib.h>
+
+// Helper function to sort projects by capital requirement
+int compare(const void *a, const void *b) {
+    int *projA = *(int **)a;
+    int *projB = *(int **)b;
+    return projA[0] - projB[0];
+}
+
+int findMaximizedCapital(int k, int w, int* profits, int profitsSize, int* capital, int capitalSize) {
+    int n = profitsSize;
+    int **projects = (int **)malloc(n * sizeof(int *));
+    for (int i = 0; i < n; ++i) {
+        projects[i] = (int *)malloc(2 * sizeof(int));
+        projects[i][0] = capital[i];
+        projects[i][1] = profits[i];
+    }
+    qsort(projects, n, sizeof(int *), compare); // Sort projects by their capital requirements
+    int idx = 0;
+    for (int i = 0; i < k; ++i) {
+        while (idx < n && projects[idx][0] <= w) {
+            w += projects[idx][1]; // Add the profit from the best project
+            ++idx;
+        }
+        if (idx == n) break;
+    }
+    for (int i = 0; i < n; ++i) {
+        free(projects[i]);
+    }
+    free(projects);
+    return w;
+}
+```
+
+## Solution in JavaScript
+
+```javascript
+class PriorityQueue {
+    constructor(comparator = (a, b) => a - b) {
+        this._heap = [];
+        this._comparator = comparator;
+    }
+
+    size() {
+        return this._heap.length;
+    }
+
+    isEmpty() {
+        return this.size() === 0;
+    }
+
+    peek() {
+        return this.isEmpty() ? undefined : this._heap[0];
+    }
+
+    push(...values) {
+        values.forEach(value => {
+            this._heap.push(value);
+            this._siftUp();
+        });
+        return this.size();
+    }
+
+    pop() {
+        const poppedValue = this.peek();
+        const bottom = this.size() - 1;
+        if (bottom > 0) {
+            this._swap(0, bottom);
+        }
+        this._heap.pop();
+        this._siftDown();
+        return poppedValue;
+    }
+
+    _parent(idx) {
+        return Math.floor((idx - 1) / 2);
+    }
+
+    _left(idx) {
+        return idx * 2 + 1;
+    }
+
+    _right(idx) {
+        return idx * 2 + 2;
+    }
+
+    _swap(i, j) {
+        [this._heap[i], this._heap[j]] = [this._heap[j], this._heap[i]];
+    }
+
+    _greater(i, j) {
+        return this._comparator(this._heap[i], this._heap[j]) < 0;
+    }
+
+    _siftUp() {
+        let node = this.size() - 1;
+        while (node > 0 && this._greater(node, this._parent(node))) {
+            this._swap(node, this._parent(node));
+            node = this._parent(node);
+        }
+    }
+
+    _siftDown() {
+        let node = 0;
+        while (
+            (this._left(node) < this.size() && this._greater(this._left(node), node)) ||
+            (this._right(node) < this.size() && this._greater(this._right(node), node))
+        ) {
+            let maxChild = (this._right(node) < this.size() && this._greater(this._right(node), this._left(node))) ? this._right(node) : this._left(node);
+            this._swap(node, maxChild);
+            node = maxChild;
+        }
+    }
+}
+
+var findMaximizedCapital = function(k, w, profits, capital) {
+    const n = profits.length;
+    const projects = [];
+    for (let i = 0; i < n; i++) {
+        projects.push([capital[i], profits[i]]);
+    }
+    projects.sort((a, b) => a[0] - b[0]); // Sort projects by their capital requirements
+    const maxHeap = new PriorityQueue((a, b) => b - a);
+    let idx = 0;
+    for (let i = 0; i < k; i++) {
+        while (idx < n && projects[idx][0] <= w) {
+            maxHeap.push(projects[idx][1]); // Max-heap for profits
+            idx++;
+        }
+        if (!maxHeap.isEmpty()) {
+            w += maxHeap.pop(); // Add the profit from the best project
+        } else {
+            break;
+        }
+    }
+    return w;
+};
+```
+
+## Step-by-Step Algorithm
+
+1. **Input Parsing**: Convert input data into appropriate data structures (arrays, lists, or vectors).
+2. **Sorting**: Sort projects based on their capital requirements.
+3. **Priority Queue Initialization**: Initialize a max-heap (priority queue) to keep track of the most profitable projects that can be started with the current capital.
+4. **Iterative Selection**:
+    - For up to `k` times, iterate through the sorted projects.
+    - Push all feasible projects (capital requirement &lt;= current capital) into the max-heap.
+    - If the max-heap is not empty, pop the project with the maximum profit and add its profit to the current capital.
+    - Break the loop if no more projects can be started.
+5. **Return Result**: Return the current capital after selecting up to `k` projects.
+
+## Conclusion
+
+The provided solutions efficiently solve the problem of selecting at most `k` projects to maximize capital using a combination of greedy strategy and priority queues. This approach ensures that the solution is optimal and runs within acceptable time limits for the given constraints.