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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 <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




// 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




<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>
Output: 
4

 

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

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