# Smallest number dividing minimum number of elements in the array | Set 2

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

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 in O(M*log(M) + N) time using sieve (M = max(arr)) will be discussed.

• First, find maximum element, M, in the array and create a frequency table freq[] of length M + 1 to store the frequency of the numbers between 1 to M.
• Iterate the array and update freq[] as freq[arr[i]]++ for each index i.
• Now, apply the sieve algorithm. Iterate between all the elements between 1 to M + 1.
• Let’s say we are iterating for a number X.
• Create a temporary variable cnt.
• For each multiple of X between X and M {X, 2X, 3X ….} update cnt as cnt = cnt + freq[kX].
• If cnt = 0 then the answer will be X else continue iterating for the next value of X.

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  ` `// in the given array ` `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 array ` `    ``int` `freq[m + 2] = { 0 }; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``freq[arr[i]]++; ` ` `  `    ``// Sieve ` `    ``for` `(``int` `i = 1; i <= m + 1; i++) { ` `        ``int` `j = i; ` `        ``int` `cnt = 0; ` `         `  `        ``// Incrementing j ` `        ``while` `(j <= m) { ` `            ``cnt += freq[j]; ` `            ``j += i; ` `        ``} ` ` `  `        ``// If no multiples of j are ` `        ``// in the array ` `        ``if` `(!cnt) ` `            ``return` `i; ` `    ``} ` ` `  `    ``return` `m + 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  ` `    ``// in the given array  ` `    ``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 array  ` `        ``int` `freq [] = ``new` `int``[m + ``2``];  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``freq[arr[i]]++;  ` `     `  `        ``// Sieve  ` `        ``for` `(``int` `i = ``1``; i <= m + ``1``; i++) ` `        ``{  ` `            ``int` `j = i;  ` `            ``int` `cnt = ``0``;  ` `             `  `            ``// Incrementing j  ` `            ``while` `(j <= m)  ` `            ``{  ` `                ``cnt += freq[j];  ` `                ``j += i;  ` `            ``}  ` `     `  `            ``// If no multiples of j are  ` `            ``// in the array  ` `            ``if` `(cnt == ``0``)  ` `                ``return` `i;  ` `        ``}  ` `        ``return` `m + ``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 ` `# in the given array ` `def` `findMin(arr, n): ` `     `  `    ``# m stores the maximum in the array ` `    ``m ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``m ``=` `max``(m, arr[i]) ` ` `  `    ``# Frequency array ` `    ``freq ``=` `[``0``] ``*` `(m ``+` `2``) ` `    ``for` `i ``in` `range``(n): ` `        ``freq[arr[i]] ``+``=` `1` ` `  `    ``# Sieve ` `    ``for` `i ``in` `range``(``1``, m ``+` `2``): ` `        ``j ``=` `i ` `        ``cnt ``=` `0` ` `  `        ``# Incrementing j ` `        ``while` `(j <``=` `m): ` `            ``cnt ``+``=` `freq[j] ` `            ``j ``+``=` `i ` ` `  `        ``# If no multiples of j are ` `        ``# in the array ` `        ``if` `(``not` `cnt): ` `            ``return` `i ` ` `  `    ``return` `m ``+` `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  ` `    ``// in the given array  ` `    ``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 array  ` `        ``int` `[]freq = ``new` `int``[m + 2];  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``freq[arr[i]]++;  ` `     `  `        ``// Sieve  ` `        ``for` `(``int` `i = 1; i <= m + 1; i++)  ` `        ``{  ` `            ``int` `j = i;  ` `            ``int` `cnt = 0;  ` `             `  `            ``// Incrementing j  ` `            ``while` `(j <= m)  ` `            ``{  ` `                ``cnt += freq[j];  ` `                ``j += i;  ` `            ``}  ` `     `  `            ``// If no multiples of j are  ` `            ``// in the array  ` `            ``if` `(cnt == 0)  ` `                ``return` `i;  ` `        ``}  ` `        ``return` `m + 1;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main ()  ` `    ``{  ` `        ``int` `[]arr = { 2, 12, 6 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``Console.WriteLine(findMin(arr, n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01  `

Output :

```5
```

Time Complexity: O(Mlog(M) + N)

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Improved By : mohit kumar 29, AnkitRai01