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 :



title-img 




                        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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