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: 3
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: 5
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 implementstion of the above approach
C++
// C++ implementation of the above approach #include <bits/stdc++.h> #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; } |
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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 |
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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 |
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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 |
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PHP
<?php // PHP implementation of the // above approach // function to find minimum operations function minOperations(&$arr, $n) { $hashTable = array(); // sort array sort($arr); // prepare hash of array for ($i = 0; $i < $n; $i++) $hashTable[$arr[$i]]++; $res = 0; for ($i = 0; $i < $n; $i++) { if ($hashTable[$arr[$i]]) { for ($j = $i; $j < $n; $j++) if ($arr[$j] % $arr[$i] == 0) $hashTable[$arr[$j]] = 0; $res++; } } return $res; } // Driver Code $arr = array(4, 6, 2, 8, 7, 21, 24, 49, 44); $n = sizeof($arr); echo minOperations($arr, $n); // This code is contributed // by Shivi_Aggarwal ?> |
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Output:
2
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