0973EasyArrayHash TableGreedySortingHeap (Priority Queue)Simulation

Rounds Needed to Clear Every Reading

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Trains the technique from

LeetCode 2357Make Array Zero by Subtracting Equal Amounts

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Readings are given as nums.

One round picks a number k above zero, no larger than the smallest reading that is still above zero, and takes k off every reading that is above zero.

Return the fewest rounds that leave every reading at zero.

Examples

Example 1

Input: nums = [14, 3, 27, 3, 9, 0]
Output: 4

The distinct readings above zero are 14, 3, 27 and 9, so four rounds are needed: taking off 3, then 6, then 5, then 13.

Example 2

Input: nums = [5, 5, 5, 5]
Output: 1

All four readings are the same, so one round of five clears them together.

Example 3

Input: nums = [0, 0, 0]
Output: 0

Everything already reads zero, so no rounds are needed.

Constraints

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

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def minimum_operations(nums: list[int]) -> int:

Java

public int minimumOperations(int[] nums)

The intended approach

Time O(n) for n readings, one pass.Space O(n) for the set of distinct readings.

A round takes the same amount off every reading still above zero, so two readings that are equal stay equal for ever and reach zero together. Meanwhile the best a round can do is take off exactly the smallest reading still standing, which clears that one and every reading equal to it. So each round clears exactly one distinct non-zero value, and the answer is simply how many distinct non-zero readings the list holds. Counting those is one pass over a set, with no simulation at all.

Pythonentry point: minimum_operations

def minimum_operations(nums):
    seen = set()
    for v in nums:
        if v > 0:
            seen.add(v)
    return len(seen)

The obvious version

Run the rounds for real: find the smallest reading still above zero, take it off every reading above zero, and count the rounds until nothing is left standing. Costs a pass per round rather than a single count.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_minimum_operations(nums):
    values = list(nums)
    rounds = 0
    while True:
        smallest = None
        for v in values:
            if v > 0 and (smallest is None or v < smallest):
                smallest = v
        if smallest is None:
            return rounds
        for i in range(len(values)):
            if values[i] > 0:
                values[i] -= smallest
        rounds += 1

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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