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

• Last Updated : 13 Jun, 2022

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 required``int` `minOperations(``int``* arr, ``int` `n)``{``    ``int` `maxi, result = 0;` `    ``// Count the frequency of each element``    ``vector<``int``> 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 code``int` `main()``{``    ``int` `arr[] = { 2, 4, 2, 4, 4, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << minOperations(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `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``[``1000001``];``        ``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 required``def` `minOperations(arr, n):` `    ``result ``=` `0``    ` `    ``# Count the frequency of each element``    ``freq ``=` `[``0``] ``*` `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 code``if` `__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 approach``using` `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``[1000001];``        ``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`

## PHP

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## Javascript

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Output:

`1`

Time Complexity: O(n + m√m), where n represents the size of the given array and m represents the maximum element in the array.

Auxiliary Space: O(1000001), no extra space is required, so it is a constant.

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