#3354

Easy

LeetCode Problem Workspace

Make Array Elements Equal to Zero

Learn how to transform an integer array to zeros using simulation and directional selection efficiently and reliably.

Problem Statement

You are given an integer array nums with at least one element equal to zero. You may start at any zero element and pick a movement direction, either left or right. From the starting position, move step by step in the chosen direction, reducing elements along the path according to the described simulation until no further valid moves remain.

Your goal is to determine the minimum number of starting positions and directions required to transform all elements of nums into zero. Return the count of valid selections that achieve this goal, considering the constraints that array length is small enough for direct simulation.

Examples

Example 1

Input: nums = [1,0,2,0,3]

Output: 2

The only possible valid selections are the following:

Example 2

Input: nums = [2,3,4,0,4,1,0]

Output: 0

There are no possible valid selections.

Constraints

1 <= nums.length <= 100
0 <= nums[i] <= 100
There is at least one element i where nums[i] == 0.

Solution Approach

Simulate Moves From Each Zero

Iterate through each element that is zero and try both left and right movements. For each direction, simulate the subtraction process until you either reach the array boundary or no elements remain positive. Count selections that lead to all zeros.

Track State Efficiently

Maintain a temporary array copy for each simulation to prevent overwriting the original. This ensures accurate evaluation of whether a starting selection and direction reduce all elements to zero without interfering with other trials.

Leverage Problem Constraints

Since nums.length <= 100, brute-force simulation is feasible. This allows testing all zero positions and both directions without optimization, ensuring correctness by exhaustive simulation rather than advanced data structures.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n^2) in the worst case since each zero element is tried in both directions over the array. Space complexity is O(n) for temporary array copies during simulation.

What Interviewers Usually Probe

Noticing the small input size implies a simulation-based solution is acceptable.
Expecting candidates to handle edge cases where no valid selection exists.
Looking for recognition that directional moves must be tested from each zero position.

Common Pitfalls or Variants

Common pitfalls

Overwriting the original array during simulation, corrupting other trials.
Failing to check both left and right directions for each zero.
Miscounting valid selections when multiple zeros interact in overlapping paths.

Follow-up variants

Count the minimum steps to reduce all elements to zero rather than the number of valid selections.
Allow subtracting variable values instead of fixed ones during each move.
Extend the array length or element range to test simulation efficiency under higher constraints.

FAQ

What is the main approach to Solve Make Array Elements Equal to Zero?

Use direct simulation from each zero element in both directions, decrementing values and checking if all elements reach zero.

Can I optimize beyond simulation for small arrays?

Since the array size is at most 100, straightforward simulation is sufficient and simple to implement correctly.

Why test both left and right directions from each zero?

Some paths reduce all elements to zero only in one direction, so checking both ensures all valid selections are counted.

How do overlapping paths affect counting valid selections?

Each starting zero and direction is evaluated independently, preventing interference from previous trials by using array copies.

Does this problem always guarantee at least one solution?

Not necessarily. There may be cases where no starting position and direction can reduce all elements to zero, yielding zero valid selections.

terminal

Solution

Solution 1: Enumeration + Prefix Sum

Suppose we initially move to the left and encounter a non-zero element. In that case, we need to decrement this element by one, then change the direction of movement and continue moving.

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class Solution:
    def countValidSelections(self, nums: List[int]) -> int:
        s = sum(nums)
        ans = l = 0
        for x in nums:
            if x:
                l += x
            elif l * 2 == s:
                ans += 2
            elif abs(l * 2 - s) == 1:
                ans += 1
        return ans

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