Minimum operations for reducing Array to 0 by subtracting smaller element from a pair repeatedly

• Difficulty Level : Medium
• Last Updated : 30 Nov, 2021

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:

1. Traverse the array and push all the elements which are greater than 0, in the priority queue.
2. Create a variable op, to store the number of operations, and initialise it with 0.
3. 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.
4. Repeat the above operation until the priority queue becomes empty.
5. Print op, as the answer to this question.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include 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 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);     cout << setElementstoZero(arr, N);     return 0;}

Java

 // Java code for the above approachimport java.util.*;class CustomComparator implements Comparator {    @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 pq            = new PriorityQueue(                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

Javascript



Output
3

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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