### 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;
time += Math.max(0, k - (time - lastSeen.get(task) - 1));
time += 1;
}
return time;
}
}```
```

```                        ```Solution in Python :

class Solution:
last_times = {}
cur_time = 0
wait_time = k - (cur_time - last_times[task]) + 1
cur_time += wait_time
cur_time += 1
return cur_time```
```

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