Patients With A Condition - Leetcode Solution

Problem Description:

You are given an array patients, where each patients[i] represents the time when the ith patient was diagnosed with a condition, and the patient needs a treatment. Each treatment takes k minutes, meaning that once a treatment is started, a patient needs k minutes to be cured. Treatments can't be started while any other treatment is ongoing (i.e., overlapping treatments are not allowed). Your task is to find the minimum number of periods of k minutes needed to treat all the patients.

Input:

patients: An array of integers representing the time when each patient was diagnosed with a condition, where 1 <= patients.length <= 10^5 and 0 <= patients[i] <= 1000.

k: A positive integer representing the length of a treatment, where 1 <= k <= 1000.

Output:

An integer representing the minimum number of periods of k minutes needed to treat all the patients.

Solution:

The problem can be solved using a greedy approach. The idea is to sort the patients' array in increasing order and then iterate through it, checking if the current patient's treatment can be done in the current period or the next period of length k.

Let's take an example to understand the approach:

patients = [1, 2, 4, 7, 8], k = 3

First sort the patients' array in increasing order: [1, 2, 4, 7, 8]

Now iterate through the patients' array, checking if the treatment can be done in the current period or the next period of length k. Initially, assign the start time of the first period as 1.

• For the first patient, treatment can be started at 1 and it will take 3 minutes. So, the end time of the first period will be 4.
• For the second patient, treatment can be started at 2, which is greater than the end time of the first period (4). So, the treatment of the second patient will be started in the next period, i.e., 4. So, the end time of the second period will be 7.
• For the third patient, treatment can be started at 4, which is greater than the end time of the second period (7). So, the treatment of the third patient will be started in the next period, i.e., 7. So, the end time of the third period will be 10.
• For the fourth patient, treatment can be started at 10, which is greater than the end time of the third period (10). So, the treatment of the fourth patient will be started in the next period, i.e., 10. So, the end time of the fourth period will be 13.
• For the fifth patient, treatment can be started at 13, which is greater than the end time of the fourth period (13). So, the treatment of the fifth patient will be started in the next period, i.e., 13. So, the end time of the fifth period will be 16.

Therefore, the minimum number of periods of k minutes needed to treat all the patients is 5.

The time complexity of the above approach is O(nlogn), where n is the length of the patients' array, due to the sorting operation. The space complexity is O(1).

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