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# Find minimum number K such that sum of array after multiplication by K exceed S

• Last Updated : 23 Mar, 2021

Given an array arr[] of N elements and an integer S, the task is to find the minimum number K such that the sum of the array elements does not exceed S after multiplying all the elements by K.
Examples:

Input: arr[] = { 1 }, S = 50
Output: 51
Explanation:
The sum of array elements is 1.
Now the multiplication of 1 with 51 gives 51 which is > 50.
Hence the minimum value of K is 51.
Input: arr[] = { 10, 7, 8, 10, 12, 19 }, S = 200
Output:
Explanation:
The sum of array elements is 66.
Now the multiplication of 66 with 4 gives 256 > 200.
Hence the minimum value of K is 4.

Approach:

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ``using` `namespace` `std;` `// Function to return the minimum value of k``// that satisfies the given condition``int` `findMinimumK(``int` `a[], ``int` `n, ``int` `S)``{``    ``// store sum of array elements``    ``int` `sum = 0;` `    ``// Calculate the sum after``    ``for` `(``int` `i = 0; i < n; i++) {``        ``sum += a[i];``    ``}` `    ``// return minimum possible K``    ``return` `ceil``(((S + 1) * 1.0)``                ``/ (sum * 1.0));``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 10, 7, 8, 10, 12, 19 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);``    ``int` `S = 200;` `    ``cout << findMinimumK(a, n, S);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;``import` `java.lang.Math;` `class` `GFG {` `// Function to return the minimum value of k``// that satisfies the given condition``static` `int` `findMinimumK(``int` `a[], ``int` `n, ``int` `S)``{``    ` `    ``// Store sum of array elements``    ``int` `sum = ``0``;` `    ``// Calculate the sum after``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``sum += a[i];``    ``}` `    ``// Return minimum possible K``    ``return` `(``int``) Math.ceil(((S + ``1``) * ``1.0``) /``                               ``(sum * ``1.0``));``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { ``10``, ``7``, ``8``, ``10``, ``12``, ``19` `};``    ``int` `n = a.length;``    ``int` `S = ``200``;``    ``System.out.print(findMinimumK(a, n, S));``}``}` `// This code is contributed by shivanisinghss2110`

## Python3

 `# Python3 implementation of the approach``import` `math` `# Function to return the minimum value of k``# that satisfies the given condition``def` `findMinimumK(a, n, S) :` `    ``# store sum of array elements``    ``sum` `=` `0` `    ``# Calculate the sum after``    ``for` `i ``in` `range``(``0``,n):``        ``sum` `+``=` `a[i]` `    ``# return minimum possible K``    ``return` `math.ceil(((S ``+` `1``) ``*` `1.0``)``/``(``sum` `*` `1.0``))` `# Driver code``a ``=` `[ ``10``, ``7``, ``8``, ``10``, ``12``, ``19` `]``n ``=` `len``(a)``s ``=` `200``print``(findMinimumK(a, n, s))` `# This code is contributed by Sanjit_Prasad`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG {` `// Function to return the minimum value of k``// that satisfies the given condition``static` `int` `findMinimumK(``int` `[]a, ``int` `n, ``int` `S)``{``    ` `    ``// Store sum of array elements``    ``int` `sum = 0;` `    ``// Calculate the sum after``    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``sum += a[i];``    ``}` `    ``// Return minimum possible K``    ``return` `(``int``) Math.Ceiling(((S + 1) * 1.0) /``                                  ``(sum * 1.0));``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]a = { 10, 7, 8, 10, 12, 19 };``    ``int` `n = a.Length;``    ``int` `S = 200;``    ` `    ``Console.Write(findMinimumK(a, n, S));``}``}` `// This code is contributed by sapnasingh4991`

## Javascript

 ``
Output:
`4`

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

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