# Make Array Zero by Subtracting Equal Amounts

Published: Sep 21, 2022

Easy Simulation Array Hash Table

## Introduction

This problem looks like a simulation required. However, if we carefully read the problem, it turns out the answer is just a number of times. Think what will go on from the problem description. The number of unique non-zero values is the answer.

## Problem Description

You are given a non-negative integer array `nums`. In one operation, you must:

• Choose a positive integer `x` such that `x` is less than or equal to the smallest non-zero element in `nums`.
• Subtract `x` from every positive element in `nums`.

Return the minimum number of operations to make every element in nums equal to 0.

Constraints:

• `1 <= nums.length <= 100`
• `0 <= nums[i] <= 100`

https://leetcode.com/problems/make-array-zero-by-subtracting-equal-amounts/

## Examples

``````Example 1
Input: nums = [1,5,0,3,5]
Output: 3
Explanation:
In the first operation, choose x = 1. Now, nums = [0,4,0,2,4].
In the second operation, choose x = 2. Now, nums = [0,2,0,0,2].
In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].
``````
``````Example 2
Input: nums = [0]
Output: 0
Explanation: Each element in nums is already 0 so no operations are needed.
``````

## Analysis

The answer is the number of unique non-zero values, so take a set at first. If 0 doesn’t exist in the set, the length of the set is the answer. Otherwise subtract 1 from the length of the set.

## Solution

``````class MakeArrayZeroBySubtractingEqualAmounts:
def minimumOperations(self, nums: List[int]) -> int:
ss = set(nums)
return len(ss) if 0 not in ss else len(ss) - 1
``````

## Complexities

• Time: `O(n)`
• Space: `O(n)`