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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2301_2400.s2365_task_scheduler_ii;

// #Medium #Array #Hash_Table #Simulation #2022_08_14_Time_70_ms_(55.56%)_Space_123.3_MB_(11.11%)

import java.util.HashMap;

/**
 * 2365 - Task Scheduler II\.
 *
 * Medium
 *
 * You are given a **0-indexed** array of positive integers `tasks`, representing tasks that need to be completed **in order** , where `tasks[i]` represents the **type** of the i^th task.
 *
 * You are also given a positive integer `space`, which represents the **minimum** number of days that must pass **after** the completion of a task before another task of the **same** type can be performed.
 *
 * Each day, until all tasks have been completed, you must either:
 *
 * *   Complete the next task from `tasks`, or
 * *   Take a break.
 *
 * Return _the **minimum** number of days needed to complete all tasks_.
 *
 * **Example 1:**
 *
 * **Input:** tasks = [1,2,1,2,3,1], space = 3
 *
 * **Output:** 9
 *
 * **Explanation:** 
 *
 * One way to complete all tasks in 9 days is as follows: 
 *
 * Day 1: Complete the 0th task. 
 *
 * Day 2: Complete the 1st task.
 *
 * Day 3: Take a break.
 *
 * Day 4: Take a break. 
 *
 * Day 5: Complete the 2nd task. 
 *
 * Day 6: Complete the 3rd task.
 *
 * Day 7: Take a break. 
 *
 * Day 8: Complete the 4th task. 
 *
 * Day 9: Complete the 5th task. 
 *
 * It can be shown that the tasks cannot be completed in less than 9 days.
 *
 * **Example 2:**
 *
 * **Input:** tasks = [5,8,8,5], space = 2
 *
 * **Output:** 6
 *
 * **Explanation:** 
 *
 * One way to complete all tasks in 6 days is as follows:
 *
 * Day 1: Complete the 0th task.
 *
 * Day 2: Complete the 1st task. 
 *
 * Day 3: Take a break.
 *
 * Day 4: Take a break.
 *
 * Day 5: Complete the 2nd task.
 *
 * Day 6: Complete the 3rd task. 
 *
 * It can be shown that the tasks cannot be completed in less than 6 days.
 *
 * **Constraints:**
 *
 * *   1 <= tasks.length <= 10⁵
 * *   1 <= tasks[i] <= 10⁹
 * *   `1 <= space <= tasks.length`
**/
public class Solution {
    public long taskSchedulerII(int[] tasks, int space) {
        long days = 0;
        space++;
        HashMap lastOccurence = new HashMap<>();
        for (int i = 0; i < tasks.length; i++) {
            if (lastOccurence.containsKey(tasks[i])) {
                long lastTimeOccurred = lastOccurence.get(tasks[i]);
                long daysDifference = days - lastTimeOccurred;
                if (daysDifference < space) {
                    days += (space - daysDifference);
                }
            }
            lastOccurence.put(tasks[i], days);
            days++;
        }
        return days;
    }
}