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# 2050. Parallel Courses III

## Problem Statement

<br>

You are given an integer `n`, which indicates that there are `n` courses labeled from `1` to `n`. You are also given a 2D integer array `relations` where `relations[j] = [prevCoursej, nextCoursej]` denotes that course `prevCoursej` has to be completed **before** course `nextCoursej` (prerequisite relationship). Furthermore, you are given a **0-indexed** integer array `time` where `time[i]` denotes how many **months** it takes to complete the `(i+1)th` course.

You must find the **minimum** number of months needed to complete all the courses following these rules:

* You may start taking a course at **any time** if the prerequisites are met.
* **Any number of courses** can be taken at the **same time**.

Return *the **minimum** number of months needed to complete all the courses*.

**Note:** The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/10/07/ex1.png)

<pre><code><strong>Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
</strong><strong>Output: 8
</strong><strong>Explanation: The figure above represents the given graph and the time required to complete each course. 
</strong>We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.
</code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/10/07/ex2.png)

<pre><code><strong>Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
</strong><strong>Output: 12
</strong><strong>Explanation: The figure above represents the given graph and the time required to complete each course.
</strong>You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.
</code></pre>

&#x20;

**Constraints:**

* `1 <= n <= 5 * 104`
* `0 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)`
* `relations[j].length == 2`
* `1 <= prevCoursej, nextCoursej <= n`
* `prevCoursej != nextCoursej`
* All the pairs `[prevCoursej, nextCoursej]` are **unique**.
* `time.length == n`
* `1 <= time[i] <= 104`
* The given graph is a directed acyclic graph.

## Intuition

```
Approach:

Use Kahn Algorithm to find the start nodes,
Apply that,

Maintain a distance array to store all the distances uptill that point and update at
each step
```

### Links

<https://leetcode.com/problems/parallel-courses-iii/description/?envType=daily-question&envId=2023-10-18>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int minimumTime(int n, vector<vector<int>>& relations, vector<int>& time) {
        vector<vector<int>> adj(n+1);
        vector<int> indeg(n+1,0);

        for(auto &it: relations) {
            indeg[it[1]]++;
            adj[it[0]].push_back(it[1]);
        }

        vector<int> dist(n+1, 0);
        queue<int> q;
        for(int i=1; i<=n; i++) {
            if(indeg[i] == 0){
                q.push(i);
                dist[i] = time[i-1];
                cout<<i<<" "<<dist[i]<<endl;
            }
        }

        while(!q.empty()){
            int temp = q.front();
            q.pop();

            for(auto &it: adj[temp]){
                dist[it] = max(dist[it], time[it-1]+dist[temp]);
                if(--indeg[it] == 0)
                    q.push(it);
            }
        }

        return *max_element(dist.begin(), dist.end());
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###


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