**Task Run - Facebook Top Interview Questions**

### Problem Statement :

You are given a list of integers tasks where each different integer represents a different task type, and a non-negative integer k. Each task takes one unit of time to complete and each task must be done in order, but you must have k units of time between doing two tasks of the same type. At any time, you can be doing a task or waiting. Return the amount of time it takes to complete all the tasks. Constraints n ≤ 100,000 where n is the length of tasks Example 1 Input tasks = [0, 1, 0, 1] k = 2 Output 5 Explanation The tasks get run the following way: [0, 1, wait, 0, 1]. Note that before running the second 0 task we must wait one unit to have 2 units of time before running it again. We can just run the second 1 task right away since it has already been 2 units of time since we last did task 1. Example 2 Input tasks = [0, 0, 1, 1] k = 1 Output 6 Explanation The tasks get run the following way: [0, wait, 0, 1, wait, 1].

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& tasks, int k) {
unordered_map<int, int> lastdone;
int curr = 0;
k += 1;
for (int t : tasks) {
if (lastdone.count(t)) {
curr = max(curr, lastdone[t] + k);
}
lastdone[t] = curr++;
}
return curr;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] tasks, int k) {
Map<Integer, Integer> lastSeen = new HashMap<>();
int time = 0;
for (int task : tasks) {
if (lastSeen.containsKey(task))
time += Math.max(0, k - (time - lastSeen.get(task) - 1));
lastSeen.put(task, time);
time += 1;
}
return time;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, tasks, k):
last_times = {}
cur_time = 0
for j, task in enumerate(tasks):
if task in last_times and cur_time - last_times[task] <= k:
wait_time = k - (cur_time - last_times[task]) + 1
cur_time += wait_time
last_times[task] = cur_time
cur_time += 1
return cur_time
```

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