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 solving 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 a 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++ 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;
} |
// 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 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# 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 |
<script> // Javascript implementation of the approach // Function to return the smallest number // that divides minimum number of elements function findMin(arr, n)
{ // m stores the maximum in the array
var m = 0;
for ( var i = 0; i < n; i++)
m = Math.max(m, arr[i]);
// Frequency table
var cnt = Array(m+2).fill(0);
// Loop to factorize
for ( var i = 0; i < n; i++) {
// sqrt factorization of the numbers
for ( var 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 ( var i = 1; i <= m + 1; i++)
if (cnt[i] == 0) {
return i;
}
return -1;
} // Driver code var arr = [2, 12, 6];
var n = arr.length;
document.write( findMin(arr, n)); </script> |
5
Time Complexity: O(N * sqrt(max(arr))).
Auxiliary Space: O(max(arr))