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Generate an N-length array having maximum element minimized and sum of array elements divisible by K
• Last Updated : 19 May, 2021

Given two positive integers N and K, the task is to minimize the maximum element of the array formed such that the sum of array elements is positive and divisible by K.

Examples:

Input: N = 4, K = 50
Output: 13
Explanation The generated array is {12, 13, 12, 13}. Sum of the array is 50, which is divisible by K (= 50). Maximum element present in the array is 13, which is minimum possible.

Input: N = 3, K = 3
Output: 1

Approach: The given problem can be solved on the basis of the following observations:

From the above observations, the minimized maximum value of the constructed array is the ceil value of (K/N).

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to minimize the maximum``// element present in an N-length array``// having sum of elements divisible by K``int` `minimumValue(``int` `N, ``int` `K)``{``    ``// Return the ceil value of (K / N)``    ``return` `ceil``((``double``)K / (``double``)N);``}` `// Driver Code``int` `main()``{``    ``int` `N = 4, K = 50;``    ``cout << minimumValue(N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Function to minimize the maximum``// element present in an N-length array``// having sum of elements divisible by K``static` `int` `minimumValue(``int` `N, ``int` `K)``{``   ` `    ``// Return the ceil value of (K / N)``    ``return``(``int``)Math.ceil((``double``)K / (``double``)N);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``4``, K = ``50``;``     ` `    ``System.out.print(minimumValue(N, K));``}``}` `// This code is contributed by code_hunt.`

## Python3

 `# Python3 program for the above approach``import` `math` `# Function to minimize the maximum``# element present in an N-length array``# having sum of elements divisible by K`  `def` `minimumValue(N, K):``    ``# Return the ceil value of (K / N)``    ``return` `math.ceil(K ``/` `N)`  `# Driver Code``N ``=` `4``K ``=` `50``print``(minimumValue(N, K))` `# This code is contributed by abhinavjain194`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to minimize the maximum``// element present in an N-length array``// having sum of elements divisible by K``static` `int` `minimumValue(``int` `N, ``int` `K)``{``  ` `    ``// Return the ceil value of (K / N)``    ``return``(``int``)Math.Ceiling((``double``)K / (``double``)N);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `N = 4, K = 50;``    ` `    ``Console.WriteLine(minimumValue(N, K));``}``}` `// This code is contributed by ukasp`

## Javascript

 ``
Output:
`13`

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

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