# Minimum operations to make Array elements 0 by decrementing pair or single element

• Difficulty Level : Medium
• Last Updated : 05 Aug, 2022

Given an array arr[] of size N, the task is to find the minimum number of operations required to reduce all three elements of the array to zero. Following operations are allowed:

• Reduce 2 different array elements by one.
• Reduce a single array element by one.

#### Example:

Input: arr[] = {1, 2, 3}, N = 3
Output: 3
Explanation : Operation 1: reduce 3 and 2 to get {1, 1, 2}
Operation 2: reduuce 1 and 2 to get {1, 0, 1}
Operation 3: reduce both 1s to get {0, 0, 0}

Input: arr[] = {5, 1, 2, 9, 8}, N = 5
Output: 13

#### Approach:

This problem can be solved using greedy approach. The idea is to reduce the 2 largest elements at a time or (if not possible) 1 at a time.  As we need the largest elements in each step, we can use a max heap.

The following steps can be taken to solve this approach:

• Initiate a count variable as 0.
• Insert all the elements in a max heap.
• Reduce the two largest elements.
• Insert the reduced values again into the heap.
• Repeat above mentioned steps until all array elements become zero and increase the count at each iteration.
• Stop when all array elements are zero.

Below is the implementation of the above approach:

## Java

 // Java program to reduce array to zero in minimum stepsimport java.util.*;  class GFG {      public static int reduceArray(int arr[], int N)    {          int count = 0;        PriorityQueue pq = new PriorityQueue<>();          for (int i = 0; i < N; i++) {            pq.add(arr[i] * -1);        }        while (pq.size() > 1) {            int temp1 = pq.poll();            int temp2 = pq.poll();            count++;            temp1++;            temp2++;            if (temp1 != 0)                pq.add(temp1);            if (temp2 != 0)                pq.add(temp2);        }        if (pq.size() > 0)            count -= pq.poll();        return count;    }      // Driver Code    public static void main(String[] args)    {        int arr[] = { 1, 2, 3 };        int N = 3;        int count = reduceArray(arr, N);        System.out.println(count);    }}

Output

3

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

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