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 = 5Input: 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 in O(M*log(M) + N) time using a sieve (M = max(arr)) will be discussed.
- First, find the 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 <bits/stdc++.h>using namespace std;// Function to return the smallest number// that divides minimum number of elements // in the given arrayint 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 codeint 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 arraydef 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 codearr = [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 |
Javascript
<script>// Javascript implementation of the approach// Function to return the smallest number// that divides minimum number of elements // in the given arrayfunction 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 array var freq = Array(m+2).fill(0); for (var i = 0; i < n; i++) freq[arr[i]]++; // Sieve for (var i = 1; i <= m + 1; i++) { var j = i; var 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 codevar arr = [2, 12, 6];var n = arr.length;document.write( findMin(arr, n));</script> |
Output
5
Time Complexity: O(Mlog(M) + N)
Auxiliary Space: O(M), where M is the maximum element in the given array.
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