# Minimum days required to cure N persons

• Last Updated : 29 Jan, 2022

Given an array arr[], representing the ages of N persons in a hospital and there is only one doctor who can give an immune dose to at most P persons per day, the task is to find the minimum count of days required to give the cure such that a high risk person and a normal person does not get a dose on the same day.
Note: Any person with age ≤ 10 or ≥ 60 is contemplated as a high risk person.

Examples:

Input: arr[] = {9, 80, 27, 72, 79}, P = 2
Output: 3
Explanation:
There are 4 high risk persons. Therefore, on days 1 and 2, 2 high risk persons can be immuned. On the last day, the normal person can be immuned. Therefore, 3 days are required. Approach: Follow the steps below to solve the problem:

1. Initialize a variable, say risk, to store the count of persons whose age is less than or equal to 10 and greater than or equal to 60.
2. Initialize a variable, say normal_risk, to store the count of persons whose age is in the range [11, 59]
3. Traverse the array and check if the age is less than or equal to 10 or greater than or equal to 60 or not. If found to be true, then increment the value of risk.
4. Otherwise, increment the value of normal_risk.
5. Finally, print the value of ceil(risk / P) + ceil(normal_risk / P)

Below is the implementation of the above approach:

## C++

 `// C++ Program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find minimum count of days required` `// to give a cure such that the high risk person` `// and risk person does not get a dose on same day.` `void` `daysToCure(``int` `arr[], ``int` `N, ``int` `P)` `{`   `    ``// Stores count of persons whose age is` `    ``// less than or equal to 10 and` `    ``// greater than or equal to 60.` `    ``int` `risk = 0;`   `    ``// Stores the count of persons` `    ``// whose age is in the range [11, 59]` `    ``int` `normal_risk = 0;`   `    ``// Traverse the array arr[]` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// If age less than or equal to 10` `        ``// or greater than or equal to 60` `        ``if` `(arr[i] >= 60 || arr[i] <= 10) {`   `            ``// Update risk` `            ``risk++;` `        ``}` `        ``else` `{`   `            ``// Update normal_risk` `            ``normal_risk++;` `        ``}` `    ``}`   `    ``// Calculate days to cure risk` `    ``// and normal_risk persons` `    ``int` `days = (risk / P) + (risk % P > 0)` `               ``+ (normal_risk / P)` `               ``+ (normal_risk % P > 0);`   `    ``// Print the days` `    ``cout << days;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given array` `    ``int` `arr[] = { 9, 80, 27, 72, 79 };`   `    ``// Size of the array` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Given P` `    ``int` `P = 2;`   `    ``daysToCure(arr, N, P);`   `    ``return` `0;` `}`

## Java

 `// Java Program for the above approach ` `class` `GFG` `{` `  ``// Function to find minimum count of days required ` `  ``// to give a cure such that the high risk person ` `  ``// and risk person does not get a dose on same day. ` `  ``static` `void` `daysToCure(``int` `arr[], ``int` `N, ``int` `P) ` `  ``{ ` ` `  `    ``// Stores count of persons whose age is ` `    ``// less than or equal to 10 and ` `    ``// greater than or equal to 60. ` `    ``int` `risk = ``0``; ` ` `  `    ``// Stores the count of persons ` `    ``// whose age is in the range [11, 59] ` `    ``int` `normal_risk = ``0``; ` ` `  `    ``// Traverse the array arr[] ` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{ ` ` `  `      ``// If age less than or equal to 10 ` `      ``// or greater than or equal to 60 ` `      ``if` `(arr[i] >= ``60` `|| arr[i] <= ``10``)` `      ``{ ` ` `  `        ``// Update risk ` `        ``risk++; ` `      ``} ` `      ``else` `      ``{ ` ` `  `        ``// Update normal_risk ` `        ``normal_risk++; ` `      ``} ` `    ``} ` ` `  `    ``// Calculate days to cure risk ` `    ``// and normal_risk persons ` `    ``int` `days = (risk / P) +  (normal_risk / P);` ` `  `    ``if``(risk % P > ``0``)` `    ``{` `      ``days++;` `    ``}` ` `  `    ``if``(normal_risk % P > ``0``)` `    ``{` `      ``days++;` `    ``}` ` `  `    ``// Print the days ` `    ``System.out.print(days); ` `  ``}  `   `    ``public` `static` `void` `main(String[] args) {` `        ``// Given array ` `        ``int` `arr[] = { ``9``, ``80``, ``27``, ``72``, ``79` `}; ` `     `  `        ``// Size of the array ` `        ``int` `N = arr.length; ` `     `  `        ``// Given P ` `        ``int` `P = ``2``;   ` `        ``daysToCure(arr, N, P);` `    ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 Program for the above approach`   `# Function to find minimum count of days required` `# to give a cure such that the high risk person` `# and risk person does not get a dose on same day.` `def` `daysToCure(arr, N, P):`   `    ``# Stores count of persons whose age is` `    ``# less than or equal to 10 and` `    ``# greater than or equal to 60.` `    ``risk ``=` `0`   `    ``# Stores the count of persons` `    ``# whose age is in the range [11, 59]` `    ``normal_risk ``=` `0`   `    ``# Traverse the array arr[]` `    ``for` `i ``in` `range``(N):`   `        ``# If age less than or equal to 10` `        ``# or greater than or equal to 60` `        ``if` `(arr[i] >``=` `60` `or` `arr[i] <``=` `10``):`   `            ``# Update risk` `            ``risk ``+``=` `1` `        ``else``:`   `            ``# Update normal_risk` `            ``normal_risk ``+``=` `1`   `    ``# Calculate days to cure risk` `    ``# and normal_risk persons` `    ``days ``=` `(risk ``/``/` `P) ``+` `(risk ``%` `P > ``0``) ``+` `(normal_risk ``/``/` `P) ``+` `(normal_risk ``%` `P > ``0``)`   `    ``# Print the days` `    ``print` `(days)`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``# Given array` `    ``arr ``=` `[``9``, ``80``, ``27``, ``72``, ``79` `]`   `    ``# Size of the array` `    ``N ``=` `len``(arr)`   `    ``# Given P` `    ``P ``=` `2`   `    ``daysToCure(arr, N, P)`   `    ``# This code is contributed by mohit kumar 29.`

## C#

 `// C# Program for the above approach ` `using` `System;` `class` `GFG ` `{`   `  ``// Function to find minimum count of days required ` `  ``// to give a cure such that the high risk person ` `  ``// and risk person does not get a dose on same day. ` `  ``static` `void` `daysToCure(``int``[] arr, ``int` `N, ``int` `P) ` `  ``{ `   `    ``// Stores count of persons whose age is ` `    ``// less than or equal to 10 and ` `    ``// greater than or equal to 60. ` `    ``int` `risk = 0; `   `    ``// Stores the count of persons ` `    ``// whose age is in the range [11, 59] ` `    ``int` `normal_risk = 0; `   `    ``// Traverse the array arr[] ` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{ `   `      ``// If age less than or equal to 10 ` `      ``// or greater than or equal to 60 ` `      ``if` `(arr[i] >= 60 || arr[i] <= 10)` `      ``{ `   `        ``// Update risk ` `        ``risk++; ` `      ``} ` `      ``else` `      ``{ `   `        ``// Update normal_risk ` `        ``normal_risk++; ` `      ``} ` `    ``} `   `    ``// Calculate days to cure risk ` `    ``// and normal_risk persons ` `    ``int` `days = (risk / P) +  (normal_risk / P);`   `    ``if``(risk % P > 0)` `    ``{` `      ``days++;` `    ``}`   `    ``if``(normal_risk % P > 0)` `    ``{` `      ``days++;` `    ``}`   `    ``// Print the days ` `    ``Console.Write(days); ` `  ``} `   `  ``// Driver code` `  ``static` `void` `Main() ` `  ``{`   `    ``// Given array ` `    ``int``[] arr = { 9, 80, 27, 72, 79 }; `   `    ``// Size of the array ` `    ``int` `N = arr.Length; `   `    ``// Given P ` `    ``int` `P = 2;   ` `    ``daysToCure(arr, N, P);` `  ``}` `}`   `// This code is contributed by divyesh072019.`

## Javascript

 ``

Output

`3`

Time Complexity: O(N)
Auxiliary Space: O(1)

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