Minimum number of given operations required to reduce the array to 0 element

• Last Updated : 14 May, 2021

Given an array arr[] of N integers. The task is to find the minimum number of given operations required to reduce the array to 0 elements. In a single operation, any element can be chosen from the array and all of its multiples get removed including itself.
Examples:

Input: arr[] = {2, 4, 6, 3, 4, 6, 8}
Output:
Operation 1: Choose 2 and delete all the multiples, arr[] = {3}
Operation 3: Choose 3 and the array gets reduced to 0 element.
Input: arr[] = {2, 4, 2, 4, 4, 4}
Output:

Naive approach: Find minimum from the array at each step and traverse the entire array to find multiples of these elements and delete them.
Efficient approach:

• Create a count array that stores the count of each number in the array.
• Since we know that for a number x the elements which satisfy the condition (A % x == 0) are actually the multiples of x and hence we need to find the multiples for every number and set their frequencies to 0 including the chosen element itself.
• Now for every number, we traverse its multiples once and subtract the value of the count of that number from all of its multiples.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach#include using namespace std; // Function to return the minimum// operations requiredint minOperations(int* arr, int n){    int maxi, result = 0;     // Count the frequency of each element    vector freq(1000001, 0);    for (int i = 0; i < n; i++) {        int x = arr[i];        freq[x]++;    }     // Maximum element from the array    maxi = *(max_element(arr, arr + n));    for (int i = 1; i <= maxi; i++) {        if (freq[i] != 0) {             // Find all the multiples of i            for (int j = i * 2; j <= maxi; j = j + i) {                 // Delete the multiples                freq[j] = 0;            }             // Increment the operations            result++;        }    }    return result;} // Driver codeint main(){    int arr[] = { 2, 4, 2, 4, 4, 4 };    int n = sizeof(arr) / sizeof(arr);     cout << minOperations(arr, n);     return 0;}

Java

 // Java implementation of the approachimport java.util.Arrays; class GFG{     // Function to return the minimum    // operations required    static int minOperations(int[] arr, int n)    {        int maxi, result = 0;         // Count the frequency of each element        int[] freq = new int;        for (int i = 0; i < n; i++)        {            int x = arr[i];            freq[x]++;        }         // Maximum element from the array        maxi = Arrays.stream(arr).max().getAsInt();        for (int i = 1; i <= maxi; i++)        {            if (freq[i] != 0)            {                 // Find all the multiples of i                for (int j = i * 2; j <= maxi; j = j + i)                {                     // Delete the multiples                    freq[j] = 0;                }                 // Increment the operations                result++;            }        }        return result;    }     // Driver code    public static void main(String[] args)    {        int arr[] = {2, 4, 2, 4, 4, 4};        int n = arr.length;         System.out.println(minOperations(arr, n));    }} // This code is contributed by 29AjayKumar

Python3

 # Python3 implementation of the approach # Function to return the minimum# operations requireddef minOperations(arr, n):     result = 0         # Count the frequency of each element    freq =  * 1000001         for i in range(0, n):        freq[arr[i]] += 1     # Maximum element from the array    maxi = max(arr)    for i in range(1, maxi+1):        if freq[i] != 0:             # Find all the multiples of i            for j in range(i * 2, maxi+1, i):                 # Delete the multiples                freq[j] = 0             # Increment the operations            result += 1             return result # Driver codeif __name__ == "__main__":     arr = [2, 4, 2, 4, 4, 4]    n = len(arr)     print(minOperations(arr, n)) # This code is contributed by Rituraj Jain

C#

 // C# implementation of above approachusing System;using System.Linq; class GFG{     // Function to return the minimum    // operations required    static int minOperations(int[] arr, int n)    {        int maxi, result = 0;         // Count the frequency of each element        int[] freq = new int;        for (int i = 0; i < n; i++)        {            int x = arr[i];            freq[x]++;        }         // Maximum element from the array        maxi = arr.Max();        for (int i = 1; i <= maxi; i++)        {            if (freq[i] != 0)            {                 // Find all the multiples of i                for (int j = i * 2; j <= maxi; j = j + i)                {                     // Delete the multiples                    freq[j] = 0;                }                 // Increment the operations                result++;            }        }        return result;    }     // Driver code    public static void Main(String[] args)    {        int []arr = {2, 4, 2, 4, 4, 4};        int n = arr.Length;         Console.WriteLine(minOperations(arr, n));    }} // This code is contributed by Rajput-Ji



Javascript


Output:
1

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