# Smallest number dividing minimum number of elements in the Array

Given an array arr[] of N integers, the task is to find the smallest number that divides the minimum number of elements from the array.

Examples:

Input: arr[] = {2, 12, 6}
Output: 5
Here, 1 divides 3 elements
2 divides 3 elements
3 divides 2 elements
4 divides 1 element
5 divides no element
6 divides 2 elements
7 divides no element
8 divides no element
9 divides no element
10 divides no element
11 divides no element
12 divides 1 element
5 is the smallest number not dividing any
number in the array. Thus, ans = 5

Input: arr[] = {1, 7, 9}
Output: 2

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

Approach: Let’s observe some details first. A number that divides zero elements already exists i.e. max(arr) + 1. Now, we just need to find the minimum number which divides zero numbers in the array.

In this article, an approach to solve this problem using square root factorization will be discussed. Each element will be factorised and a frequency array cnt[] of length max(arr) + 2 will be maintained to store the count of number of elements in the array, for each element between 1 to max(arr) + 1.

• For each i, factorize arr[i].
• For each factor Fij of arr[i], update cnt[Fij] as cnt[Fij]++.
• Find the smallest number k in the frequency array cnt[] with cnt[k] = 0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the smallest number ` `// that divides minimum number of elements ` `int` `findMin(``int``* arr, ``int` `n) ` `{ ` `    ``// m stores the maximum in the array ` `    ``int` `m = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``m = max(m, arr[i]); ` ` `  `    ``// Frequency table ` `    ``int` `cnt[m + 2] = { 0 }; ` ` `  `    ``// Loop to factorize ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// sqrt factorization of the numbers ` `        ``for` `(``int` `j = 1; j * j <= arr[i]; j++) { ` `            ``if` `(arr[i] % j == 0) { ` `                ``if` `(j * j == arr[i]) ` `                    ``cnt[j]++; ` `                ``else` `                    ``cnt[j]++, cnt[arr[i] / j]++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Finding the smallest number ` `    ``// with zero multiples ` `    ``for` `(``int` `i = 1; i <= m + 1; i++) ` `        ``if` `(cnt[i] == 0) { ` `            ``return` `i; ` `        ``} ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 12, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << findMin(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` ` `  `    ``// Function to return the smallest number  ` `    ``// that divides minimum number of elements  ` `    ``static` `int` `findMin(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``// m stores the maximum in the array  ` `        ``int` `m = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``m = Math.max(m, arr[i]);  ` `     `  `        ``// Frequency table  ` `        ``int` `cnt[] = ``new` `int``[m + ``2``];  ` `     `  `        ``// Loop to factorize  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `     `  `            ``// sqrt factorization of the numbers  ` `            ``for` `(``int` `j = ``1``; j * j <= arr[i]; j++)  ` `            ``{  ` `                ``if` `(arr[i] % j == ``0``) ` `                ``{  ` `                    ``if` `(j * j == arr[i])  ` `                        ``cnt[j]++;  ` `                    ``else` `                    ``{ ` `                        ``cnt[j]++; ` `                        ``cnt[arr[i] / j]++;  ` `                    ``} ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``// Finding the smallest number  ` `        ``// with zero multiples  ` `        ``for` `(``int` `i = ``1``; i <= m + ``1``; i++)  ` `            ``if` `(cnt[i] == ``0``)  ` `            ``{  ` `                ``return` `i;  ` `            ``}  ` `        ``return` `-``1``;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `arr[] = { ``2``, ``12``, ``6` `};  ` `        ``int` `n = arr.length;  ` `     `  `        ``System.out.println(findMin(arr, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the smallest number ` `# that divides minimum number of elements ` `def` `findMin(arr, n): ` `     `  `    ``# m stores the maximum in the array ` `    ``m ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``m ``=` `max``(m, arr[i]) ` ` `  `    ``# Frequency table ` `    ``cnt ``=` `[``0``] ``*` `(m ``+` `2``) ` ` `  `    ``# Loop to factorize ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# sqrt factorization of the numbers ` `        ``j ``=` `1` `        ``while` `j ``*` `j <``=` `arr[i]: ` ` `  `            ``if` `(arr[i] ``%` `j ``=``=` `0``): ` `                ``if` `(j ``*` `j ``=``=` `arr[i]): ` `                    ``cnt[j] ``+``=` `1` `                ``else``: ` `                    ``cnt[j] ``+``=` `1` `                    ``cnt[arr[i] ``/``/` `j] ``+``=` `1` `            ``j ``+``=` `1`     ` `  `    ``# Finding the smallest number ` `    ``# with zero multiples ` `    ``for` `i ``in` `range``(``1``, m ``+` `2``): ` `        ``if` `(cnt[i] ``=``=` `0``): ` `            ``return` `i ` ` `  `    ``return` `-``1` ` `  `# Driver code ` `arr ``=` `[``2``, ``12``, ``6``] ` `n ``=` `len``(arr) ` ` `  `print``(findMin(arr, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to return the smallest number  ` `    ``// that divides minimum number of elements  ` `    ``static` `int` `findMin(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``// m stores the maximum in the array  ` `        ``int` `m = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``m = Math.Max(m, arr[i]);  ` `     `  `        ``// Frequency table  ` `        ``int` `[]cnt = ``new` `int``[m + 2];  ` `     `  `        ``// Loop to factorize  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `     `  `            ``// sqrt factorization of the numbers  ` `            ``for` `(``int` `j = 1; j * j <= arr[i]; j++)  ` `            ``{  ` `                ``if` `(arr[i] % j == 0) ` `                ``{  ` `                    ``if` `(j * j == arr[i])  ` `                        ``cnt[j]++;  ` `                    ``else` `                    ``{ ` `                        ``cnt[j]++; ` `                        ``cnt[arr[i] / j]++;  ` `                    ``} ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``// Finding the smallest number  ` `        ``// with zero multiples  ` `        ``for` `(``int` `i = 1; i <= m + 1; i++)  ` `            ``if` `(cnt[i] == 0)  ` `            ``{  ` `                ``return` `i;  ` `            ``}  ` `        ``return` `-1;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{  ` `        ``int` `[]arr = { 2, 12, 6 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``Console.WriteLine(findMin(arr, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```5
```

Time Complexity: O(N * sqrt(max(arr))).

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