### 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```
```

## Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

## Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

## Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

## The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

## Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever