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

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: 4 
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: 
 

• Find the sum of all elements of the array, store it in a variable sum.
• Take the ceil division of (S + 1) with sum. This will be the required minimum value of K.

Below is the implementation of the above approach:
 

// C++ implementation of the approach

#include <bits/stdc++.h>
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 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 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# 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

<script>

    // Javascript implementation of the approach 
    
    // Function to return the minimum value of k
    // that satisfies the given condition
    function findMinimumK(a, n, S)
    {
        // store sum of array elements
        let sum = 0;

        // Calculate the sum after
        for (let i = 0; i < n; i++) {
            sum += a[i];
        }

        // return minimum possible K
        return Math.ceil(((S + 1) * 1.0) / (sum * 1.0));
    }
    
    let a = [ 10, 7, 8, 10, 12, 19 ];
    let n = a.length;
    let S = 200;
  
    document.write(findMinimumK(a, n, S));

</script>

4

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