Minimize removal from Array so that in any pair one element is multiple of other

Last Updated : 07 Apr, 2022

Given an array arr[] of size N, the task is to count the minimum number of elements required to be removed from the given array such that whenever any pair (arr[i], arr[j]) is picked, where i != j and 0 ≤ i < j < N, either arr[i] is multiple of arr[j] or vice versa.

Examples:

Input: N = 5, arr[] = {4, 3, 4, 5, 2}
Output: 2
Explanation: Currently, pair (arr[2], arr[3]) does not satisfy given condition.
Similarly, with i = 4, and i =5 as well.
Remove arr[2]= 3, and arr[4] = 5, to would make array satisfy given conditions.
Therefore minimum removal of elements is 2. And array is {4, 4, 2}.

Input: N = 3, arr[] = {2, 2, 4}
Output: 0
Explanation: As, array already satisfies the given condition, there's no need to remove any element.

Approach: The problem can be solved by using Dynamic Programming and Sieve of Eratosthenes based on the following idea:

Get the frequencies of the array elements. Now find the element having maximum number of multiples present in the array using the frequency table and dynamic programming. All the remaining elements should be deleted to achieve the condition using minimum deletions.

Follow the steps mentioned below to solve the problem:

Traverse the arr[] and store the maximum element.
Store the frequency of the array elements.
Traversing through dp array and store frequency of ith element in dp[i].
- Then use Sieve of Eratosthenes to traverse through multiples of i.
- Update the current dp value with the maximum of dp[i] and current dp value.
Finding the max element in dp[] array after doing operations

Below is the implementation of the above approach:

C++

// C++ code for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to do required operation
int solve(int N, vector<int>& arr)
{
    // Initializing maxi to store
    // max element in given array
    int maxi = INT_MIN;

    // Finding out max element in
    // given array arr
    for (int i = 0; i < N; i++) {
        maxi = max(maxi, arr[i]);
    }

    // Initializing a vector
    // to store frequency of array elements
    vector<int> freq(maxi + 1);

    // Traversing from 0 to N
    // to store frequency of array elements
    for (int i = 0; i < N; i++) {
        freq[arr[i]]++;
    }

    // Initializing dp vector
    vector<int> dp(maxi + 1);

    // Initializing final answer
    // to store minimum steps
    int answer = 0;

    // Traversing through dp array
    for (int i = 1; i <= maxi; i++) {

        // Storing frequency of i'th
        // element in dp[i]
        dp[i] += freq[i];

        // Using sieve of Eratosthenes to
        // traverse through multiples
        // of i
        for (int j = 2 * i; j <= maxi;
             j += i) {

            // Updating dp[j] as discussed
            // in approach with max of dp[j]
            // and occurrence of i
            dp[j] = max(dp[j], dp[i]);
        }
    }

    // Finding the max element in
    // dp vector after doing operations
    int max_in_dp
        = *max_element(dp.begin(), dp.end());

    // Printing the final answer
    return (N - max_in_dp);
}

// Driver Code
int main()
{
    // Taking input
    int N = 5;
    vector<int> arr = { 4, 3, 4, 5, 2 };

    // Function call
    cout << solve(N, arr);
    return 0;
}

Java

// Java code for the above approach
import java.util.*;

class GFG {

  // Function to do required operation
  static int solve(int N, int[] arr)
  {
    
    // Initializing maxi to store
    // max element in given array
    int maxi = Integer.MIN_VALUE;

    // Finding out max element in
    // given array arr
    for (int i = 0; i < N; i++) {
      maxi = Math.max(maxi, arr[i]);
    }

    // Initializing a vector
    // to store frequency of array elements
    int[] freq = new int[maxi + 1];

    // Traversing from 0 to N
    // to store frequency of array elements
    for (int i = 0; i < N; i++) {
      freq[arr[i]]++;
    }

    // Initializing dp vector
    int[] dp = new int[maxi + 1];

    // Initializing final answer
    // to store minimum steps
    int answer = 0;

    // Traversing through dp array
    for (int i = 1; i <= maxi; i++) {

      // Storing frequency of i'th
      // element in dp[i]
      dp[i] += freq[i];

      // Using sieve of Eratosthenes to
      // traverse through multiples
      // of i
      for (int j = 2 * i; j <= maxi;
           j += i) {

        // Updating dp[j] as discussed
        // in approach with max of dp[j]
        // and occurrence of i
        dp[j] = Math.max(dp[j], dp[i]);
      }
    }

    // Finding the max element in
    // dp vector after doing operations
    int max_in_dp
      = Arrays.stream(dp).max().getAsInt();

    // Printing the final answer
    return (N - max_in_dp);
  }

  // Driver Code
  public static void main (String[] args)
  {
    
    // Taking input
    int N = 5;
    int arr[] = { 4, 3, 4, 5, 2 };

    // Function call
    System.out.print(solve(N, arr));
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach

# Function to do required operation
def solve(N, arr):

    # Initializing maxi to store
    # max element in given array
    maxi = -9999999;

    # Finding out max element in
    # given array arr
    for i in range(N):
        maxi = max(maxi, arr[i]);
    
    # Initializing a vector
    # to store frequency of array elements
    freq = [0]*(maxi+1);

    # Traversing from 0 to N
    # to store frequency of array elements
    for i in range(N):
        freq[arr[i]] = freq[arr[i]] + 1;
    
    # Initializing final answer
    # to store minimum steps
    answer = 0;
    dp = [0]*(maxi + 1);
    
    # Traversing through dp array
    for i in range(1, maxi):

        # Storing frequency of i'th
        # element in dp[i]
        dp[i] = dp[i]+  freq[i];

        # Using sieve of Eratosthenes to
        # traverse through multiples
        # of i
        for j in range(2 * i,maxi, i):
            
            # Updating dp[j] as discussed
            # in approach with max of dp[j]
            # and occurrence of i
            dp[j] = max(dp[j], dp[i]);
        
    # Finding the max element in
    # dp vector after doing operations
    max_in_dp = max(dp);

    # Printing the final answer
    return (N - max_in_dp);

# Driver Code

    # Taking input
N = 5;
arr = [4, 3, 4, 5, 2];

    # Function call
print(solve(N, arr));
  
# This code is contributed by Potta Lokesh

// C# implementation of above approach
using System;
using System.Collections.Generic;
using System.Linq;

public class GFG{

  // Function to do required operation
  static int solve(int N, int[] arr)
  {

    // Initializing maxi to store
    // max element in given array
    int maxi = Int32.MinValue;

    // Finding out max element in
    // given array arr
    for (int i = 0; i < N; i++) {
      maxi = Math.Max(maxi, arr[i]);
    }

    // Initializing a vector
    // to store frequency of array elements
    int[] freq = new int[maxi + 1];

    // Traversing from 0 to N
    // to store frequency of array elements
    for (int i = 0; i < N; i++) {
      freq[arr[i]]++;
    }

    // Initializing dp vector
    int[] dp = new int[maxi + 1];

    // Initializing final answer
    // to store minimum steps
    int answer = 0;

    // Traversing through dp array
    for (int i = 1; i <= maxi; i++) {

      // Storing frequency of i'th
      // element in dp[i]
      dp[i] += freq[i];

      // Using sieve of Eratosthenes to
      // traverse through multiples
      // of i
      for (int j = 2 * i; j <= maxi;
           j += i) {

        // Updating dp[j] as discussed
        // in approach with max of dp[j]
        // and occurrence of i
        dp[j] = Math.Max(dp[j], dp[i]);
      }
    }

    // Finding the max element in
    // dp vector after doing operations
    int max_in_dp
      = dp.Max();

    // Printing the final answer
    return (N - max_in_dp);
  }

  // Driver Code
  static public void Main (){

    // Taking input
    int N = 5;
    int[] arr = { 4, 3, 4, 5, 2 };

    // Function call
    Console.Write(solve(N, arr));
  }
}

// This code is contributed by code_hunt.

JavaScript

    <script>
        // JavaScript code for the above approach

        const INT_MIN = -2147483647 - 1;
        const INT_MAX = 2147483647;

        // Function to do required operation
        const solve = (N, arr) => {
            // Initializing maxi to store
            // max element in given array
            let maxi = INT_MIN;

            // Finding out max element in
            // given array arr
            for (let i = 0; i < N; i++) {
                maxi = Math.max(maxi, arr[i]);
            }

            // Initializing a vector
            // to store frequency of array elements
            let freq = new Array(maxi + 1).fill(0);

            // Traversing from 0 to N
            // to store frequency of array elements
            for (let i = 0; i < N; i++) {
                freq[arr[i]]++;
            }

            // Initializing dp vector
            let dp = new Array(maxi + 1).fill(0);

            // Initializing final answer
            // to store minimum steps
            let answer = 0;

            // Traversing through dp array
            for (let i = 1; i <= maxi; i++) {

                // Storing frequency of i'th
                // element in dp[i]
                dp[i] += freq[i];

                // Using sieve of Eratosthenes to
                // traverse through multiples
                // of i
                for (let j = 2 * i; j <= maxi;
                    j += i) {

                    // Updating dp[j] as discussed
                    // in approach with max of dp[j]
                    // and occurrence of i
                    dp[j] = Math.max(dp[j], dp[i]);
                }
            }

            // Finding the max element in
            // dp vector after doing operations
            let max_in_dp = Math.max(...dp);

            // Printing the final answer
            return (N - max_in_dp);
        }

        // Driver Code

        // Taking input
        let N = 5;
        let arr = [4, 3, 4, 5, 2];

        // Function call
        document.write(solve(N, arr));

    // This code is contributed by rakeshsahni

    </script>

Output

Time Complexity: O(M*log(M)), where M is maximum element present in given array
Auxiliary Space: O(M)

Closest pair in an Array such that one number is multiple of the other

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Minimize removal from Array so that in any pair one element is multiple of other

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