Given an array arr[] of size N, the task is to find the minimum number of operations required to make all array elements zero. In one operation, select a pair of elements and subtract the smaller element from both elements in the array.
Example:
Input: arr[] = {1, 2, 3, 4}
Output: 3
Explanation: Pick the elements in the following sequence:
Operation 1: Pick elements at indices {3, 2}: arr[]={1, 2, 0, 1}
Operation 2: Pick elements at indices {1, 3}: arr[]={1, 1, 0, 0}
Operation 3: Pick elements at indices {2, 1}: arr[]={0, 0, 0, 0}Input: arr[] = {2, 2, 2, 2}
Output: 2
Approach: This problem can be solved using a priority queue. To solve the below problem, follow the below steps:
- Traverse the array and push all the elements which are greater than 0, in the priority queue.
- Create a variable op, to store the number of operations, and initialise it with 0.
- Now, iterate over the priority queue pq till its size is greater than one in each iteration:
- Increment the value of variable op.
- Then select the top two elements, let’s say p and q to apply the given operation.
- After applying the operation, one element will definitely become 0. Push the other one back into the priority queue if it is greater than zero.
- Repeat the above operation until the priority queue becomes empty.
- Print op, as the answer to this question.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum number// of operations required to make all// array elements zeroint setElementstoZero(int arr[], int N){ // Create a priority queue priority_queue<int> pq; // Variable to store the number // of operations int op = 0; for (int i = 0; i < N; i++) { if (arr[i] > 0) { pq.push(arr[i]); } } // Iterate over the priority queue // till size is greater than 1 while (pq.size() > 1) { // Increment op by 1 op += 1; auto p = pq.top(); pq.pop(); auto q = pq.top(); pq.pop(); // If the element is still greater // than zero again push it again in pq if (p - q > 0) { pq.push(p); } } // Return op as the answer return op;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4 }; int N = sizeof(arr) / sizeof(arr[0]); cout << setElementstoZero(arr, N); return 0;} |
Java
// Java code for the above approachimport java.util.*;class CustomComparator implements Comparator<Integer> { @Override public int compare(Integer number1, Integer number2) { int value = number1.compareTo(number2); // elements are sorted in reverse order if (value > 0) { return -1; } else if (value < 0) { return 1; } else { return 0; } }}class GFG { // Function to find the minimum number // of operations required to make all // array elements zero static int setElementstoZero(int arr[], int N) { // Create a priority queue PriorityQueue<Integer> pq = new PriorityQueue<Integer>( new CustomComparator()); // Variable to store the number // of operations int op = 0; for (int i = 0; i < N; i++) { if (arr[i] > 0) { pq.add(arr[i]); } } // Iterate over the priority queue // till size is greater than 1 while (pq.size() > 1) { // Increment op by 1 op = op + 1; Integer p = pq.poll(); Integer q = pq.poll(); // If the element is still greater // than zero again push it again in pq if (p - q > 0) { pq.add(p); } } // Return op as the answer return op; } // Driver Code public static void main(String[] args) { int arr[] = { 1, 2, 3, 4 }; int N = arr.length; System.out.println(setElementstoZero(arr, N)); }}// This code is contributed by Potta Lokesh |
Python3
# Python program for the above approach# Function to find the minimum number# of operations required to make all# array elements zerodef setElementstoZero(arr, N): # Create a priority queue pq = [] # Variable to store the number # of operations op = 0 for i in range(N): if (arr[i] > 0): pq.append(arr[i]) pq.sort() # Iterate over the priority queue # till size is greater than 1 while (len(pq) > 1): # Increment op by 1 op += 1 p = pq[len(pq) - 1] pq.pop() q = pq[len(pq)-1] pq.pop() # If the element is still greater # than zero again push it again in pq if (p - q > 0): pq.append(p) pq.sort() # Return op as the answer return op# Driver Codearr = [1, 2, 3, 4]N = len(arr)print(setElementstoZero(arr, N))# This code is contributed by Saurabh Jaiswal |
C#
// C# code for the above approachusing System;using System.Collections.Generic;public class GFG { // Function to find the minimum number // of operations required to make all // array elements zero static int setElementstoZero(int[] arr, int N) { // Create a priority queue List<int> pq = new List<int>(); // Variable to store the number // of operations int op = 0; for (int i = 0; i < N; i++) { if (arr[i] > 0) { pq.Add(arr[i]); } } // Iterate over the priority queue // till size is greater than 1 while (pq.Count > 1) { pq.Sort(); pq.Reverse(); // Increment op by 1 op = op + 1; int p = pq[0]; int q = pq[1]; pq.RemoveRange(0, 2); // If the element is still greater // than zero again push it again in pq if (p - q > 0) { pq.Add(p); } } // Return op as the answer return op; } // Driver Code public static void Main(String[] args) { int[] arr = { 1, 2, 3, 4 }; int N = arr.Length; Console.WriteLine(setElementstoZero(arr, N)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum number// of operations required to make all// array elements zerofunction setElementstoZero(arr, N){ // Create a priority queue var pq = []; // Variable to store the number // of operations var op = 0; for(var i = 0; i < N; i++) { if (arr[i] > 0) { pq.push(arr[i]); } } pq.sort((a,b) => a-b); // Iterate over the priority queue // till size is greater than 1 while (pq.length > 1) { // Increment op by 1 op += 1; var p = pq[pq.length-1]; pq.pop(); var q = pq[pq.length-1]; pq.pop(); // If the element is still greater // than zero again push it again in pq if (p - q > 0) { pq.push(p); } pq.sort((a,b) => a-b); } // Return op as the answer return op;}// Driver Codevar arr = [ 1, 2, 3, 4 ];var N = arr.length;document.write(setElementstoZero(arr, N));// This code is contributed by rutvik_56.</script> |
Output
3
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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