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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

5

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

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