Improved task 2895

Author: javadev

src/main/java/g2801_2900/s2895_minimum_processing_time/readme.md

@@ -32,12 +32,18 @@ Hence, it can be shown that the minimum time taken to execute all the tasks is 1
 
 **Explanation:**
 
-It's optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20. Time taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18. Time taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23. Hence, it can be shown that the minimum time taken to execute all the tasks is 23. 
+It's optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20.
+
+Time taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.
+
+Time taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.
+
+Hence, it can be shown that the minimum time taken to execute all the tasks is 23. 
 
 **Constraints:**
 
 *   `1 <= n == processorTime.length <= 25000`
 *   <code>1 <= tasks.length <= 10<sup>5</sup></code>
 *   <code>0 <= processorTime[i] <= 10<sup>9</sup></code>
 *   <code>1 <= tasks[i] <= 10<sup>9</sup></code>
-*   `tasks.length == 4 * n`
+*   `tasks.length == 4 * n`