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# Minimum number of operations required to delete all elements of the array

Given an integer array arr, the task is to print the minimum number of operations required to delete all elements of the array.
In an operation, any element from the array can be chosen at random and every element divisible by it can be removed from the array.

Examples:

Input: arr[] = {2, 4, 6, 3, 5, 10}
Output:
Choosing 2 as the first element will remove 2, 4, 6 and 10 from the array.
Now choose 3 which will result in the removal of 3.
Finally, the only element left to choose is 5.

Input: arr[] = {2, 5, 3, 7, 11}
Output:

Approach: For optimal results, the smallest element from the array should be chosen from the remaining elements one after another until all the elements of the array are deleted.

1. Sort the array in ascending order and find the multiple of element in complete vector.
2. For each element which are divisible by choose element mark it 0, and decrease the value of N.
3. Return the value of N.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `int` `solve(``int` `N, vector<``int``> A)``{``    ``sort(A.begin(), A.end());``    ``int` `a = 0;``    ``for` `(``int` `i = 0; i < A.size() - 1; i++) {``        ``int` `c = 0;``        ``if` `(A[i] != 0) {``            ``for` `(``int` `j = i + 1; j < A.size(); j++) {``                ``if` `(A[j] % A[i] == 0 && A[j] != 0``                    ``&& A[i] != 0) {``                    ``A[j] = 0;``                    ``N--;``                ``}``            ``}``        ``}``    ``}``    ``if` `(N == 0) {``        ``N = 1;``    ``}``    ``return` `N;``}``// Driver program``int` `main()``{``    ``vector<``int``> v = { 4, 6, 2, 8, 7, 21, 24, 49, 44 };``    ``int` `n = v.size();` `    ``cout << solve(n, v);``    ``return` `0;``}``// This code is contributed by Raunak_Kumar`

## Java

 `/*package whatever //do not write package name here */``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `  ``// Java implementation of the above approach` `  ``static` `int` `solve(``int` `N, ``int``[] A)``  ``{``    ``Arrays.sort(A);``    ``int` `a = ``0``;``    ``for` `(``int` `i = ``0``; i < A.length - ``1``; i++) {``      ``int` `c = ``0``;``      ``if` `(A[i] != ``0``) {``        ``for` `(``int` `j = i + ``1``; j < A.length; j++) {``          ``if` `(A[j] % A[i] == ``0` `&& A[j] != ``0``              ``&& A[i] != ``0``) {``            ``A[j] = ``0``;``            ``N--;``          ``}``        ``}``      ``}``    ``}``    ``if` `(N == ``0``) {``      ``N = ``1``;``    ``}``    ``return` `N;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String args[])``  ``{``    ``int``[] v = { ``4``, ``6``, ``2``, ``8``, ``7``, ``21``, ``24``, ``49``, ``44` `};``    ``int` `n = v.length;` `    ``System.out.println(solve(n, v));``  ``}``}` `// This code is contributed by shinjanpatra`

## Python3

 `# Python implementation of the above approach``def` `solve(N,A):``    ``A.sort()``    ``a ``=` `0``    ``for` `i ``in` `range``(``len``(A) ``-` `1``):``        ``c ``=` `0``        ``if` `(A[i] !``=` `0``):``            ``for` `j ``in` `range``(i ``+` `1``,``len``(A)):``                ``if` `(A[j] ``%` `A[i] ``=``=` `0` `and` `A[j] !``=` `0` `and` `A[i] !``=` `0``):``                    ``A[j] ``=` `0``                    ``N ``-``=` `1``    ``if` `(N ``=``=` `0``):``        ``N ``=` `1``    ``return` `N` `# Driver program``v ``=` `[ ``4``, ``6``, ``2``, ``8``, ``7``, ``21``, ``24``, ``49``, ``44` `]``n ``=` `len``(v)` `print``(solve(n, v))` `# This code is contributed by shinjanpatra`

## C#

 `// C# code to implement the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `  ``static` `int` `solve(``int` `N, ``int``[] A)``  ``{``    ``Array.Sort(A);``//    int a = 0;``    ``for` `(``int` `i = 0; i < A.Length - 1; i++) {``//      int c = 0;``      ``if` `(A[i] != 0) {``        ``for` `(``int` `j = i + 1; j < A.Length; j++) {``          ``if` `(A[j] % A[i] == 0 && A[j] != 0``              ``&& A[i] != 0) {``            ``A[j] = 0;``            ``N--;``          ``}``        ``}``      ``}``    ``}``    ``if` `(N == 0) {``      ``N = 1;``    ``}``    ``return` `N;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] v = { 4, 6, 2, 8, 7, 21, 24, 49, 44 };``    ``int` `n = v.Length;` `    ``Console.WriteLine(solve(n, v));``  ``}``}` `// This code is contributed by phasing17`

## Javascript

 ``

Output

`2`

Time Complexity: O(N2 logN), where N is the size of the array
Space Complexity: O(1)

Approach: For optimal results, the smallest element from the array should be chosen from the remaining elements one after another until all the elements of the array are deleted.

• Sort the array in ascending order and prepare a hash for occurrences.
• For each unmarked element starting from beginning mark all elements which are divisible by choose element, and increase the result counter.

Below is the implementation of the above approach

## C++

 `// C++ implementation of the above approach` `#include ``#define MAX 10000` `using` `namespace` `std;` `int` `hashTable[MAX];` `// function to find minimum operations``int` `minOperations(``int` `arr[], ``int` `n)``{``    ``// sort array``    ``sort(arr, arr + n);` `    ``// prepare hash of array``    ``for` `(``int` `i = 0; i < n; i++)``        ``hashTable[arr[i]]++;` `    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(hashTable[arr[i]]) {``            ``for` `(``int` `j = i; j < n; j++)``                ``if` `(arr[j] % arr[i] == 0)``                    ``hashTable[arr[j]] = 0;``            ``res++;``        ``}``    ``}` `    ``return` `res;``}` `// Driver program``int` `main()``{``    ``int` `arr[] = { 4, 6, 2, 8, 7, 21, 24, 49, 44 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << minOperations(arr, n);``    ``return` `0;``}`

## Java

 `//Java implementation of the above approach``import` `java.util.*;``class` `Solution``{``static` `final` `int` `MAX=``10000``;` `static` `int` `hashTable[]= ``new` `int``[MAX];` `// function to find minimum operations``static` `int` `minOperations(``int` `arr[], ``int` `n)``{``    ``// sort array``    ``Arrays.sort(arr);` `    ``// prepare hash of array``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``hashTable[arr[i]]++;` `    ``int` `res = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``if` `(hashTable[arr[i]]!=``0``) {``            ``for` `(``int` `j = i; j < n; j++)``                ``if` `(arr[j] % arr[i] == ``0``)``                    ``hashTable[arr[j]] = ``0``;``            ``res++;``        ``}``    ``}` `    ``return` `res;``}` `// Driver program``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``4``, ``6``, ``2``, ``8``, ``7``, ``21``, ``24``, ``49``, ``44` `};``    ``int` `n = arr.length;` `    ``System.out.print( minOperations(arr, n));`` ` `}``}``// This code is contributed by Arnab Kundu`

## Python 3

 `# Python 3 implementation of``# the above approach``MAX` `=` `10000` `hashTable ``=` `[``0``] ``*` `MAX` `# function to find minimum operations``def` `minOperations(arr, n):``    ` `    ``# sort array``    ``arr.sort()` `    ``# prepare hash of array``    ``for` `i ``in` `range``(n):``        ``hashTable[arr[i]] ``+``=` `1` `    ``res ``=` `0``    ``for` `i ``in` `range``(n) :``        ``if` `(hashTable[arr[i]]) :``            ``for` `j ``in` `range``(i, n):``                ``if` `(arr[j] ``%` `arr[i] ``=``=` `0``):``                    ``hashTable[arr[j]] ``=` `0``            ``res ``+``=` `1` `    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[ ``4``, ``6``, ``2``, ``8``, ``7``, ``21``, ``24``, ``49``, ``44` `]``    ``n ``=` `len``(arr)` `    ``print``(minOperations(arr, n))` `# This code is contributed``# by ChitraNayal`

## C#

 `using` `System;` `// C# implementation of the above approach ``public` `class` `Solution``{``public` `const` `int` `MAX = 10000;` `public` `static` `int``[] hashTable = ``new` `int``[MAX];` `// function to find minimum operations ``public` `static` `int` `minOperations(``int``[] arr, ``int` `n)``{``    ``// sort array ``    ``Array.Sort(arr);` `    ``// prepare hash of array ``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``hashTable[arr[i]]++;``    ``}` `    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(hashTable[arr[i]] != 0)``        ``{``            ``for` `(``int` `j = i; j < n; j++)``            ``{``                ``if` `(arr[j] % arr[i] == 0)``                ``{``                    ``hashTable[arr[j]] = 0;``                ``}``            ``}``            ``res++;``        ``}``    ``}` `    ``return` `res;``}` `// Driver program ``public` `static` `void` `Main(``string``[] args)``{``    ``int``[] arr = ``new` `int``[] {4, 6, 2, 8, 7, 21, 24, 49, 44};``    ``int` `n = arr.Length;` `    ``Console.Write(minOperations(arr, n));` `}``}` `// This code is contributed by Shrikant13`

## PHP

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O(N logN), where N is the size of the array
Space Complexity: O(MAX)

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