Minimum number of given operations required to reduce the array to 0 element
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 element. 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: 2
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: 1
Naive approach: Find minimum from the array at each step and traverse the entire array to find multiples of this elements and delete them.
Efficient approach:
- Create a count array which stores the count of each number in the array.
- Since we know that for a number x the elements which satisfies 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 it’s multiples once and subtract the value of the count of that number from all of it’s multiples.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> 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; } |
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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 |
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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 |
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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 |
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PHP
<?php // PHP implementation of the approach // Function to return the minimum // operations required function minOperations($arr, $n) { $result = 0; $freq = array(); // Count the frequency of each element for($i = 0; $i < $n; $i++) { $freq[$arr[$i]] = 0; } for ($i = 0; $i < $n; $i++) { $x = $arr[$i]; $freq[$x]++; } // Maximum element from the array $maxi = max($arr); for ($i = 1; $i <= $maxi; $i++) { if ($freq[$i] != 0) { // Find all the multiples of i for ($j = $i * 2; $j <= $maxi; $j = $j + $i) { // Delete the multiples $freq[$j] = 0; } // Increment the operations $result++; } } return $result; } // Driver code $arr = array( 2, 4, 2, 4, 4, 4 ); $n = count($arr); echo minOperations($arr, $n); // This code is contributed by AnkitRai01 ?> |
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Output:
1
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