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; } |
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
<?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 ?> |
2
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Minimum delete operations to make all elements of array same
- Minimum operations required to make all Array elements divisible by K
- Minimum count of elements required to obtain the given Array by repeated mirror operations
- Minimum number of given operations required to reduce the array to 0 element
- Minimum number of steps required to obtain the given Array by the given operations
- Minimum number of operations required to reduce N to 1
- Minimum number of operations required to reduce N to 0
- Minimum number of operations required to obtain a given Binary String
- Minimum number of operations required to maximize the Binary String
- Find minimum changes required in an array for it to contain k distinct elements
- Minimum steps required to reduce all the elements of the array to zero
- Make all the array elements odd with minimum operations of given type
- Minimum operations to make all elements equal using the second array
- Minimum concatenation required to get strictly LIS for array with repetitive elements | Set-2
- Minimum Decrements on Subarrays required to reduce all Array elements to zero
- Minimum number of given operations required to convert a permutation into an identity permutation
- Minimum Bitwise AND operations to make any two array elements equal
- Minimum Bitwise XOR operations to make any two array elements equal
- Minimum number of segments required such that each segment has distinct elements
- Minimum number of colors required to color a Circular Array
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
