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Count of elements not divisible by any other elements of Array

  • Difficulty Level : Medium
  • Last Updated : 24 May, 2021

Given an array arr[], the task is to determine the number of elements of the array which are not divisible by any other element in the given array. 
Examples: 

Input: arr[] = {86, 45, 18, 4, 8, 28, 19, 33, 2} 
Output:
Explanation: 
The elements are {2, 19, 33, 45} are not divisible by any other array element.

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Input: arr[] = {3, 3, 3} 
Output:
 



Naive Approach: The naive approach is to iterate over the entire array and count the number of elements that are not divisible by any other elements in the given array and print the count.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
 
// Function to count the number of
// elements of array which are not
// divisible by any other element
// in the array arr[]
int count(int a[], int n)
{
    int countElements = 0;
 
    // Iterate over the array
    for (int i = 0; i < n; i++) {
 
        bool flag = true;
        for (int j = 0; j < n; j++) {
 
            // Check if the element
            // is itself or not
            if (i == j)
                continue;
 
            // Check for divisibility
            if (a[i] % a[j] == 0) {
                flag = false;
                break;
            }
        }
 
        if (flag == true)
            ++countElements;
    }
 
    // Return the final result
    return countElements;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 86, 45, 18, 4, 8,
                  28, 19, 33, 2 };
    int n = sizeof(arr) / sizeof(int);
 
    // Function Call
    cout << count(arr, n);
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to count the number of
// elements of array which are not
// divisible by any other element
// in the array arr[]
static int count(int a[], int n)
{
    int countElements = 0;
 
    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
        boolean flag = true;
        for(int j = 0; j < n; j++)
        {
 
            // Check if the element
            // is itself or not
            if (i == j)
                continue;
 
            // Check for divisibility
            if (a[i] % a[j] == 0)
            {
                flag = false;
                break;
            }
        }
 
        if (flag == true)
            ++countElements;
    }
 
    // Return the final result
    return countElements;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 86, 45, 18, 4, 8,
                  28, 19, 33, 2 };
                   
    int n = arr.length;
 
    // Function call
    System.out.print(count(arr, n));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program for
# the above approach
 
# Function to count the number of
# elements of array which are not
# divisible by any other element
# in the array arr[]
def count(a, n):
 
    countElements = 0
 
    # Iterate over the array
    for i in range (n):
 
        flag = True
        for j in range (n):
 
            # Check if the element
            # is itself or not
            if (i == j):
                continue
 
            # Check for divisibility
            if (a[i] % a[j] == 0):
                flag = False
                break
          
        if (flag == True):
            countElements += 1
    
    # Return the final result
    return countElements
 
# Driver Code
if __name__ == "__main__":
   
    # Given array
    arr = [86, 45, 18, 4,
           8, 28, 19, 33, 2]
    n = len(arr)
 
    # Function Call
    print( count(arr, n))
 
# This code is contributed by Chitranayal

C#




// C# program for the
// above approach
using System;
class GFG{
 
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element in the array []arr
static int count(int []a,
                 int n)
{
  int countElements = 0;
 
  // Iterate over the array
  for(int i = 0; i < n; i++)
  {
    bool flag = true;
    for(int j = 0; j < n; j++)
    {
      // Check if the element
      // is itself or not
      if (i == j)
        continue;
 
      // Check for divisibility
      if (a[i] % a[j] == 0)
      {
        flag = false;
        break;
      }
    }
 
    if (flag == true)
      ++countElements;
  }
 
  // Return the readonly result
  return countElements;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array
  int []arr = {86, 45, 18, 4, 8,
               28, 19, 33, 2};
 
  int n = arr.Length;
 
  // Function call
  Console.Write(count(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// Javascript program for the above approach
 
// Function to count the number of
// elements of array which are not
// divisible by any other element
// in the array arr[]
function count(a, n)
{
    let countElements = 0;
 
    // Iterate over the array
    for (let i = 0; i < n; i++) {
 
        let flag = true;
        for (let j = 0; j < n; j++) {
 
            // Check if the element
            // is itself or not
            if (i == j)
                continue;
 
            // Check for divisibility
            if (a[i] % a[j] == 0) {
                flag = false;
                break;
            }
        }
 
        if (flag == true)
            ++countElements;
    }
 
    // Return the final result
    return countElements;
}
 
// Driver Code
    // Given array
    let arr = [ 86, 45, 18, 4, 8,
                  28, 19, 33, 2 ];
    let n = arr.length;
 
    // Function Call
    document.write(count(arr, n));
 
</script>
Output
4

Time Complexity: O(N2) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, we will use the concept of Sieve of Eratosthenes. Below are the steps: 

  1. Initialize a boolean array(say v[]) of size equal to the maximum element present in the array + 1 with true at every index.
  2. Traverse the given array arr[] and change the value at the index of multiple of current elements as false in the array v[].
  3. Create a Hashmap and store the frequency of each element in it.
  4. For each element(say current_element) in the array, if v[current_element] is true then that element is not divisible by any other element in the given array and increment the count for the current element.
  5. Print the final value of count after the above steps.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// elements of array which are not
// divisible by any other element
// of same array
int countEle(int a[], int n)
{
    // Length for boolean array
    int len = 0;
 
    // Hash map for storing the
    // element and it's frequency
    unordered_map<int, int> hmap;
    for (int i = 0; i < n; i++) {
 
        // Update the maximum element
        len = max(len, a[i]);
        hmap[a[i]]++;
    }
 
    // Boolean array of size
    // of the max element + 1
    bool v[len + 1];
 
    for (int i = 0; i <= len; i++) {
        v[i] = true;
    }
 
    // Marking the multiples as false
    for (int i = 0; i < n; i++) {
 
        if (v[a[i]] == false)
            continue;
 
        for (int j = 2 * a[i];
             j <= len; j += a[i]) {
            v[j] = false;
        }
    }
 
    // To store the final count
    int count = 0;
 
    // Traverse boolean array
    for (int i = 1; i <= len; i++) {
 
        // Check if i is not divisible by
        // any other array elements and
        // appears in the array only once
        if (v[i] == true
            && hmap.count(i) == 1
            && hmap[i] == 1) {
            count += 1;
        }
    }
 
    // Return the final Count
    return count;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 86, 45, 18, 4, 8,
                  28, 19, 33, 2 };
    int n = sizeof(arr) / sizeof(int);
 
    // Function Call
    cout << countEle(arr, n);
 
    return 0;
}

Java




// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element of same array
static int countEle(int a[],
                    int n)
{
  // Length for boolean array
  int len = 0;
 
  // Hash map for storing the
  // element and it's frequency
  HashMap<Integer,
          Integer> hmap = new HashMap<>();
   
  for (int i = 0; i < n; i++)
  {
    // Update the maximum element
    len = Math.max(len, a[i]);
     
    if(hmap.containsKey(a[i]))
    {
      hmap.put(a[i],
      hmap.get(a[i]) + 1);
    }
    else
    {
      hmap.put(a[i], 1);
    }
  }
 
  // Boolean array of size
  // of the max element + 1
  boolean []v = new boolean[len + 1];
 
  for (int i = 0; i <= len; i++)
  {
    v[i] = true;
  }
 
  // Marking the multiples
  // as false
  for (int i = 0; i < n; i++)
  {
    if (v[a[i]] == false)
      continue;
 
    for (int j = 2 * a[i];
             j <= len; j += a[i])
    {
      v[j] = false;
    }
  }
 
  // To store the
  // final count
  int count = 0;
 
  // Traverse boolean array
  for (int i = 1; i <= len; i++)
  {
    // Check if i is not divisible by
    // any other array elements and
    // appears in the array only once
    if (v[i] == true &&
        hmap.containsKey(i) &&
        hmap.get(i) == 1 &&
        hmap.get(i) == 1)
    {
      count += 1;
    }
  }
 
  // Return the final
  // Count
  return count;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array
  int arr[] = {86, 45, 18, 4, 8,
               28, 19, 33, 2};
  int n = arr.length;
 
  // Function Call
  System.out.print(countEle(arr, n));
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 program for the above approach
 
# Function to count the number of
# elements of array which are not
# divisible by any other element
# of same array
def countEle(a, n):
     
    # Length for boolean array
    len = 0
 
    # Hash map for storing the
    # element and it's frequency
    hmap = {}
    for i in range(n):
         
        # Update the maximum element
        len = max(len, a[i])
        hmap[a[i]] = hmap.get(a[i], 0) + 1
 
    # Boolean array of size
    # of the max element + 1
    v = [True for i in range(len + 1)]
 
    # Marking the multiples as false
    for i in range(n):
 
        if (v[a[i]] == False):
            continue
 
        for j in range(2 * a[i], len + 1, a[i]):
            v[j] = False
 
    # To store the final count
    count = 0
 
    # Traverse boolean array
    for i in range(1, len + 1):
 
        # Check if i is not divisible by
        # any other array elements and
        # appears in the array only once
        if (v[i] == True and (i in hmap) and hmap[i] == 1):
            count += 1
 
    # Return the final Count
    return count
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [86, 45, 18, 4, 8, 28, 19, 33, 2]
    n = len(arr)
 
    # Function Call
    print(countEle(arr, n))
 
    # This code is contributed by mohit kumar 29

C#




// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element of same array
static int countEle(int []a,
                    int n)
{
  // Length for bool array
  int len = 0;
 
  // Hash map for storing the
  // element and it's frequency
  Dictionary<int,
             int> hmap =
             new Dictionary<int,
                            int>();
   
  for (int i = 0; i < n; i++)
  {
    // Update the maximum
    // element
    len = Math.Max(len, a[i]);
     
    if(hmap.ContainsKey(a[i]))
    {
      hmap[a[i]]++;
    }
    else
    {
      hmap.Add(a[i], 1);
    }
  }
 
  // Boolean array of size
  // of the max element + 1
  bool []v = new bool[len + 1];
 
  for (int i = 0; i <= len; i++)
  {
    v[i] = true;
  }
 
  // Marking the multiples
  // as false
  for (int i = 0; i < n; i++)
  {
    if (v[a[i]] == false)
      continue;
 
    for (int j = 2 * a[i];
             j <= len; j += a[i])
    {
      v[j] = false;
    }
  }
 
  // To store the
  // readonly count
  int count = 0;
 
  // Traverse bool array
  for (int i = 1; i <= len; i++)
  {
    // Check if i is not divisible by
    // any other array elements and
    // appears in the array only once
    if (v[i] == true &&
        hmap.ContainsKey(i) &&
        hmap[i] == 1 &&
        hmap[i] == 1)
    {
      count += 1;
    }
  }
 
  // Return the readonly
  // Count
  return count;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array
  int []arr = {86, 45, 18, 4, 8,
               28, 19, 33, 2};
  int n = arr.Length;
 
  // Function Call
  Console.Write(countEle(arr, n));
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
 
// Javascript program for
// the above approach
 
// Function to count the
// number of elements of
// array which are not
// divisible by any other
// element of same array
function countEle(a, n)
{
  // Length for boolean array
  let len = 0;
  
  // Hash map for storing the
  // element and it's frequency
  let hmap = new Map();
    
  for (let i = 0; i < n; i++)
  {
    // Update the maximum element
    len = Math.max(len, a[i]);
      
    if(hmap.has(a[i]))
    {
      hmap.set(a[i],
      hmap.get(a[i]) + 1);
    }
    else
    {
      hmap.set(a[i], 1);
    }
  }
  
  // Boolean array of size
  // of the max element + 1
  let v = Array.from({length: len + 1}, (_, i) => 0);
  
  for (let i = 0; i <= len; i++)
  {
    v[i] = true;
  }
  
  // Marking the multiples
  // as false
  for (let i = 0; i < n; i++)
  {
    if (v[a[i]] == false)
      continue;
  
    for (let j = 2 * a[i];
             j <= len; j += a[i])
    {
      v[j] = false;
    }
  }
  
  // To store the
  // final count
  let count = 0;
  
  // Traverse boolean array
  for (let i = 1; i <= len; i++)
  {
    // Check if i is not divisible by
    // any other array elements and
    // appears in the array only once
    if (v[i] == true &&
        hmap.has(i) &&
        hmap.get(i) == 1 &&
        hmap.get(i) == 1)
    {
      count += 1;
    }
  }
  
  // Return the final
  // Count
  return count;
}
 
 
// Driver code
 
     // Given array
  let arr = [86, 45, 18, 4, 8,
               28, 19, 33, 2];
  let n = arr.length;
  
  // Function Call
  document.write(countEle(arr, n));
 
// This code is contributed by souravghosh0416.
</script>
Output
4

Time Complexity: O(N*log(M)) where N is the number of elements in the given array and M is the maximum element in the given array. 
Auxiliary Space: O(M + N) where N is the number of elements in the given array and M is the maximum element in the given array.




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