# Minimum element whose n-th power is greater than product of an array of size n

Given an array of n integers. Find minimum x which is to be assigned to every array element such that product of all elements of this new array is strictly greater than product of all elements of the initial array.

Examples:

```Input: 4 2 1 10 6
Output: 4

Explanation: Product of elements of initial
array 4*2*1*10*6 = 480. If x = 4 then 4*4*
4*4*4 = 480, if x = 3 then 3*3*3*3*3=243.
So minimal element = 4

Input: 3 2 1 4
Output: 3

Explanation: Product of elements of initial
array 3*2*1*4 = 24. If x = 3 then 3*3*3*3
= 81, if x = 2 then 2*2*2*2 = 243. So minimal
element = 3.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach: A simple approach is to run a loop from 1 till we find the product is greater then the initial array product.
Time Complexity : O(x^n) and if used pow function then O(x * log n)

Mathematical Approach:

```Let, x^n = a1 * a2 * a3 * a4 *....* an
we have been given n and value of a1, a2, a3, ..., an.
Now take log on both sides with base e

n*logex > loge(a1) + loge(a2) +......+ loge(an)
Lets sum = loge(a1) + loge(a2) + ...... + loge(an)
n*loge x > sum
loge x > sum/n
Then take antilog on both side
x > e^(sum/n)
```

Below is the implementation of above approach.

 `// CPP program to find minimum element whose n-th ` `// power is greater than product of an array of  ` `// size n ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the minimum element ` `int` `findMin(``int` `a[], ``int` `n) ` `{ ` `    ``// loop to traverse and store the sum of log ` `    ``double` `sum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``sum += ``log``(a[i]); ``// computes sum ` ` `  `    ``// calculates the elements according to formula. ` `    ``int` `x = ``exp``(sum / n); ` ` `  `    ``// returns the minimal element ` `    ``return` `x + 1; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``// initialised array ` `    ``int` `a[] = { 3, 2, 1, 4 }; ` ` `  `    ``// computes the size of array ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``// prints out the minimal element ` `    ``cout << findMin(a, n); ` `} `

 `// JAVA Code to find Minimum element whose ` `// n-th power is greater than product of ` `// an array of size n ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``// function to find the minimum element ` `    ``static` `int` `findMin(``int` `a[], ``int` `n) ` `    ``{ ` `        ``// loop to traverse and store the  ` `        ``// sum of log ` `        ``double` `sum = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `             `  `            ``// computes sum ` `            ``sum += Math.log(a[i]);  ` `      `  `        ``// calculates the elements  ` `        ``// according to formula. ` `        ``int` `x = (``int``)Math.exp(sum / n); ` `      `  `        ``// returns the minimal element ` `        ``return` `x + ``1``; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``// initialised array ` `        ``int` `a[] = { ``3``, ``2``, ``1``, ``4` `}; ` `      `  `        ``// computes the size of array ` `        ``int` `n = a.length; ` `      `  `        ``// prints out the minimal element ` `        ``System.out.println(findMin(a, n)); ` `         `  `    ``} ` `} ` ` `  `// This code is contributed by Arnav Kr. Mandal.     `

 `# Python3 program to find minimum element ` `# whose n-th power is greater than product ` `# of an array of size n ` `import` `math as m ` ` `  `# function to find the minimum element ` `def` `findMin( a, n): ` `     `  `    ``# loop to traverse and store the ` `    ``# sum of log ` `    ``_sum ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``_sum ``+``=` `m.log(a[i]) ``# computes sum ` `     `  `    ``# calculates the elements  ` `    ``# according to formula. ` `    ``x ``=` `m.exp(_sum ``/` `n) ` `     `  `    ``# returns the minimal element ` `    ``return` `int``(x ``+` `1``) ` `     `  `# Driver program to test above function ` ` `  `# initialised array ` `a ``=` `[ ``3``, ``2``, ``1``, ``4` `] ` ` `  `# computes the size of array ` `n ``=` `len``(a) ` ` `  `# prints out the minimal element ` `print``(findMin(a, n)) ` ` `  `# This code is contributed by "Abhishek Sharma 44" `

 `// C# Code to find Minimum element whose ` `// n-th power is greater than product of ` `// an array of size n ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// function to find the minimum element ` `    ``static` `int` `findMin(``int` `[]a, ``int` `n) ` `    ``{ ` `         `  `        ``// loop to traverse and store the  ` `        ``// sum of log ` `        ``double` `sum = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `             `  `            ``// computes sum ` `            ``sum += Math.Log(a[i]);  ` `     `  `        ``// calculates the elements  ` `        ``// according to formula. ` `        ``int` `x = (``int``)Math.Exp(sum / n); ` `     `  `        ``// returns the minimal element ` `        ``return` `x + 1; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main() ` `    ``{ ` `         `  `        ``// initialised array ` `        ``int` `[]a = { 3, 2, 1, 4 }; ` `     `  `        ``// computes the size of array ` `        ``int` `n = a.Length; ` `     `  `        ``// prints out the minimal element ` `        ``Console.WriteLine(findMin(a, n)); ` `         `  `    ``} ` `} ` ` `  `// This code is contributed by vt_m.  `

 ` `

Output:
```3
```

Time Complexity: O(n * log(logn))
Auxiliary Space: O(1)