Print array elements that are divisible by at-least one other

Given an array of length N that contains only integers, the task is to print the special numbers of array. A number in this array is called Special number if it is divisible by at least one other number in the array.

Examples :

Input : 1 2 3 
Output : 2 3 
Explanation : both 2 and 3 are divisible by 1.

Input : 2 3 4 6 8 9 
Output : 4 6 8 9 
Explanation : 2 and 3 are not divisible by any other element. Rest of the element are divisible by at-least 1 element. 6 is divisible by both 2 and 3, 4 divisible by 2, 8 divisible by 2 and 4 both, 9 divisible by 3.

Input : 3 5 7 11 
Output :
Explanation : all elements are relatively prime so no special number.

A simple solution is to traverse through all elements, then check for every element if it is divisible by any other. Time complexity of this solution is O(n²)

Another solution that works better when there are many elements with not very big values. Store all array elements into hash and find out the max element in array then up-to max element find out the multiples of a given number then if multiple of array element is in hash then that number is divisible by at-least one element of array . To remove duplicate values we store the value into set because if array has 2, 3, and 6 then only 6 is divisible by at-least one element of array, both 2 and 3 divides 6 so 6 will be stored only one time. 

Implementation:

C++

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

// Function to find special numbers
void divisibilityCheck(int arr[], int n)
{

    // Storing all array elements in a hash
    // and finding maximum element in the array
    unordered_set<int> s;
    int max_ele = INT_MIN;
    for (int i = 0; i < n; i++) {
        s.insert(arr[i]);

        // Update the maximum element of the array
        max_ele = max(max_ele, arr[i]);
    }

    // Traversing the array elements and storing the array
    // multiples that are present in s in res
    unordered_set<int> res;
    for (int i = 0; i < n; i++) {

        // Check for non-zero values only
        if (arr[i] != 0) {

            // Checking the factors of current element
            for (int j = arr[i] * 2; j <= max_ele; j += arr[i]) {

                // If current factor is already part
                // of the array then store it
                if (s.find(j) != s.end())
                    res.insert(j);
            }
        }
    }

    // For non-distinct elmments
    // To store the frequency of elements
    unordered_map<int, int> mp;
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;

    unordered_map<int, int>::iterator it;
    vector<int> ans;
    for (it = mp.begin(); it != mp.end(); it++) {

        // If frequency is at least 2
        if (it->second >= 2) {
            if (res.find(it->first) == res.end()) {

                // If frequency is greater than 1 and
                // the number is not divisible by
                // any other number
                int val = it->second;

                // Then we push the element number of
                // times it is present in the vector
                while (val--)
                    ans.push_back(it->first);
            }
        }

        // If frequency is greater than 1 and the number
        // is divisible by any other number
        if (res.find(it->first) != res.end()) {
            int val = it->second;

            // Then we push the element number of
            // times it is present in the vector
            while (val--)
                ans.push_back(it->first);
        }
    }

    // Print the elements that are divisible by
    // at least one other element from the array
    for (auto x : ans)
        cout << x << " ";
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 8, 6, 9, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);

    divisibilityCheck(arr, n);

    return 0;
}

Java

// Java program to find special
// numbers in an array
import java.io.*;
import java.util.*;

class GFG {
    // Function to find
    // special numbers
    static void divisibilityCheck(List<Integer> arr,
                                  int n)
    {
        // Storing all array elements
        // in a hash and finding maximum
        // element in array
        List<Integer> s = new ArrayList<Integer>();
        int max_ele = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            s.add(arr.get(i));

            // finding maximum
            // element of array
            max_ele = Math.max(max_ele,
                               arr.get(i));
        }

        // traversing array element and
        // storing the array multiples
        // that are present in s in res.
        LinkedHashSet<Integer> res = new LinkedHashSet<Integer>();
        for (int i = 0; i < n; i++) {

            // Check for non-zero values only
            if (arr.get(i) != 0)

                // checking the factor
                // of current element
                for (int j = arr.get(i) * 2;
                     j <= max_ele;
                     j += arr.get(i)) {

                    // if factor is already
                    // part of array element
                    // then store it
                    if (s.contains(j))
                        res.add(j);
                }
        }

        // displaying elements that
        // are divisible by at least
        // one other in array
        List<Integer> list = new ArrayList<Integer>(res);
        Collections.reverse(list);

        for (Integer temp : list)
            System.out.print(temp + " ");
    }

    // Driver Code
    public static void main(String args[])
    {
        List<Integer> arr = Arrays.asList(2, 3, 8, 6, 9, 10);
        int n = arr.size();
        divisibilityCheck(arr, n);
    }
}

// This code is contributed by
// Manish Shaw(manishshaw1)

Python3

# Python3 program to find special numbers
# in an array
import math as mt

# Function to find special numbers
def divisibilityCheck(arr, n):

    # Storing all array elements in a hash
    # and finding maximum element in array
    s = dict()
    max_ele = -10**9
    for i in range(n):
        s[arr[i]] = 1

        # finding maximum element of array
        max_ele = max(max_ele, arr[i])

    # traversing array element and storing
    # the array multiples that are present
    # in s in res.
    res = dict()
    for i in range(n):

        # Check for non-zero values only
        if (arr[i] != 0):

            # checking the factor of current element
            for j in range(arr[i] * 2,
                           max_ele + 1, arr[i]):

                # if factor is already part of
                # array element then store it
                if (j in s.keys()):
                    res[j] = 1

    # displaying elements that are divisible
    # by at least one other in array
    for x in res:
        print(x, end = " ")

# Driver code
arr = [ 2, 3, 8, 6, 9, 10]
n = len(arr)
divisibilityCheck(arr, n)

# This code is contributed by
# Mohit Kumar 29

C#

// C# program to find special
// numbers in an array
using System;
using System.Linq;
using System.Collections.Generic;

class GFG {
    // Function to find
    // special numbers
    static void divisibilityCheck(List<int> arr,
                                  int n)
    {
        // Storing all array elements
        // in a hash and finding maximum
        // element in array
        List<int> s = new List<int>();
        int max_ele = Int32.MinValue;
        for (int i = 0; i < n; i++) {
            s.Add(arr[i]);

            // finding maximum element of array
            max_ele = Math.Max(max_ele,
                               arr[i]);
        }

        // traversing array element and
        // storing the array multiples
        // that are present in s in res.
        HashSet<int> res = new HashSet<int>();
        for (int i = 0; i < n; i++) {

            // Check for non-zero values only
            if (arr[i] != 0)

                // checking the factor
                // of current element
                for (int j = arr[i] * 2; j <= max_ele;
                     j += arr[i]) {

                    // if factor is already part
                    // of array element then store it
                    if (s.Contains(j))
                        res.Add(j);
                }
        }
        // displaying elements that
        // are divisible by at least
        // one other in array
        foreach(int i in res.Reverse())
            Console.Write(i + " ");
    }

    // Driver Code
    static void Main()
    {
        List<int> arr = new List<int>() { 2, 3, 8,
                                              6, 9, 10 };
        int n = arr.Count;
        divisibilityCheck(arr, n);
    }
}

// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript

<script>

// JavaScript implementation of the approach


// Function to find special numbers
function divisibilityCheck(arr, n) {

    // Storing all array elements in a hash
    // and finding maximum element in the array
    let s = new Set();
    let max_ele = Number.MIN_SAFE_INTEGER;
    for (let i = 0; i < n; i++) {
        s.add(arr[i]);

        // Update the maximum element of the array
        max_ele = Math.max(max_ele, arr[i]);
    }

    // Traversing the array elements and storing the array
    // multiples that are present in s in res
    let res = new Set();
    for (let i = 0; i < n; i++) {

        // Check for non-zero values only
        if (arr[i] != 0) {

            // Checking the factors of current element
            for (let j = arr[i] * 2; j <= max_ele; j += arr[i]) {

                // If current factor is already part
                // of the array then store it
                if (s.has(j))
                    res.add(j);
            }
        }
    }

    // For non-distinct elmments
    // To store the frequency of elements
    let mp = new Map();
    for (let i = 0; i < n; i++) {
        if (mp.has(arr[i])) {
            mp.set(arr[i], mp.get(arr[i]) + 1)
        } else {
            mp.set(arr[i], 1)
        }
    }


    let ans = [];
    for (let it of mp) {

        // If frequency is at least 2
        if (it[1] >= 2) {
            if (res.has(it[0])) {

                // If frequency is greater than 1 and
                // the number is not divisible by
                // any other number
                let val = it[1];

                // Then we push the element number of
                // times it is present in the vector
                while (val--)
                    ans.push(it[0]);
            }
        }

        // If frequency is greater than 1 and the number
        // is divisible by any other number
        if (res.has(it[0])) {
            let val = it[1];

            // Then we push the element number of
            // times it is present in the vector
            while (val--)
                ans.push(it[0]);
        }
    }

    // Print the elements that are divisible by
    // at least one other element from the array
    for (let x of ans.sort((a, b) => a - b))
        document.write(x + " ");
}

// Driver code

let arr = [2, 3, 8, 6, 9, 10];
let n = arr.length

divisibilityCheck(arr, n);

</script>

Output:

6 8 9 10

Complexity Analysis:

• Time Complexity: O(n x m), where n is the size of array and m is the maximum element in array
• Auxiliary Space: O(n)

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi.

Note: If we need results to be printed in sorted order, we can use set in place of unordered_set.