Make Array Elements Equal to Zero

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Arrays

You are given an integer array nums.

Start by selecting a starting position curr such that nums[curr] == 0, and choose a movement direction of either left or right.

After that, you repeat the following process:

If curr is out of the range [0, n - 1], this process ends.
If nums[curr] == 0, move in the current direction by incrementing curr if you are moving right, or decrementing curr if you are moving left.
Else if nums[curr] > 0:
- Decrement nums[curr] by 1.
- Reverse your movement direction (left becomes right and vice versa).
- Take a step in your new direction.

A selection of the initial position curr and movement direction is considered valid if every element in nums becomes 0 by the end of the process.

Return the number of possible valid selections.

Example 1:

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

Output: 2

Explanation:

The only possible valid selections are the following:

Choose curr = 3, and a movement direction to the left.
- [1,0,2,0,3] -> [1,0,2,0,3] -> [1,0,1,0,3] -> [1,0,1,0,3] -> [1,0,1,0,2] -> [1,0,1,0,2] -> [1,0,0,0,2] -> [1,0,0,0,2] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,1] -> [0,0,0,0,0].
Choose curr = 3, and a movement direction to the right.
- [1,0,2,0,3] -> [1,0,2,0,3] -> [1,0,2,0,2] -> [1,0,2,0,2] -> [1,0,1,0,2] -> [1,0,1,0,2] -> [1,0,1,0,1] -> [1,0,1,0,1] -> [1,0,0,0,1] -> [1,0,0,0,1] -> [1,0,0,0,0] -> [1,0,0,0,0] -> [1,0,0,0,0] -> [1,0,0,0,0] -> [0,0,0,0,0].

Example 2:

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

Output: 0

Explanation:

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

For each potential starting configuration (a zero's index and a direction), should the simulation be run on a fresh copy of the original `nums` array?
Is the process guaranteed to terminate by the pointer going out of bounds, or is it possible for it to get into an infinite loop under some circumstances?
If the input array `nums` already consists of all zeros, should every possible starting selection be considered valid?
The constraints suggest a direct simulation for each of the possible starting selections might be feasible. Is it safe to assume this approach meets the performance requirements?
To confirm the counting logic, if `nums[i]` and `nums[j]` are both 0 for `i != j`, are `(start=i, direction=right)` and `(start=j, direction=right)` considered two distinct selections?

Brute Force Solution

Approach

Imagine this is a puzzle where your only move is to subtract one from a group of `k` neighboring numbers. The brute force approach is to try every single possible sequence of moves, like exploring every path in a maze, to see if any path leads to a state where all numbers are zero.

Here's how the algorithm would work step-by-step:

Start with the original list of numbers.
Look at the list and find all the different groups of `k` neighboring numbers you could possibly operate on.
Pick one of these groups and subtract one from each number in it. This gives you a new list of numbers.
Now, with this new list, you repeat the process: find all possible groups to operate on and pick one to create yet another new list.
You keep following these chains of choices, one after another, generating new lists along the way.
If a choice ever makes a number go below zero, that particular chain of moves is a dead end. You must go back and try a different choice from a previous step.
If you ever follow a chain of moves that results in a list where every single number is zero, you've found a solution and you can stop.
If you try every single possible chain of moves and none of them lead to the all-zero list, then you know it's impossible.

Code Implementation

def make_array_zero_brute_force(numbers, window_size):
    def explore_paths(current_numbers):
        # If all numbers are zero, we have found a valid sequence of operations and exited the maze.
        if all(element == 0 for element in current_numbers):
            return True

        # Explore every possible move from the current state by selecting a starting position for the operation.
        for start_index in range(len(current_numbers) - window_size + 1):
            is_possible_move = all(current_numbers[index] > 0 for index in range(start_index, start_index + window_size))

            if is_possible_move:
                # Apply the operation to a copy so we can backtrack to the original state if this path fails.
                next_state_numbers = list(current_numbers)
                for index_in_window in range(window_size):
                    next_state_numbers[start_index + index_in_window] -= 1

                if explore_paths(next_state_numbers):
                    return True

        return False

    return explore_paths(list(numbers))

Big(O) Analysis

Time Complexity

O(n^S) where S is the sum of the numbers in the array – This approach explores a decision tree of all possible move sequences. The branching factor of this tree is n-k+1 (approximately n), as there are that many choices for which subarray to decrement at each step. The maximum depth of the tree is proportional to S, the sum of the initial array elements, since each move reduces the total sum. Therefore, the total operations approximate the number of branches raised to the power of the maximum depth, resulting in a time complexity of O(n^S).

Space Complexity

O(N * S) – The brute-force method explores different sequences of moves, implying a recursive approach. The "chains of choices" described correspond to the recursion call stack, whose maximum depth is bounded by S, the initial sum of all numbers, since each move decreases this sum. According to the description, each recursive call generates a "new list of numbers," which requires creating a copy of the array of size N. Therefore, the total auxiliary space is the stack depth (S) multiplied by the space for the array copy (N), leading to a complexity of O(N * S).

Optimal Solution

Approach

The optimal strategy avoids guesswork by processing the numbers in a single pass from left to right. At each step, the required action is forced: we must perform just enough operations to zero out the current number. By tracking the ripple effect of these forced decisions, we can efficiently determine if the entire sequence can be cleared.

Here's how the algorithm would work step-by-step:

We'll go through the numbers one by one, from start to finish. Let's keep track of a 'reduction force' that we carry along, which starts at zero.
At each number, we first see what its value becomes after being reduced by the current 'reduction force'.
If the number becomes negative at any point, the puzzle is impossible, and we can stop.
If the number is still positive, say its value is now 'V', we have no choice but to start 'V' new reduction operations at this very spot.
These 'V' new operations add to our 'reduction force' for the numbers that come next.
However, this force isn't permanent. An operation started at one spot only influences a fixed number of subsequent items, a block of size 'k'.
This means that as we move forward, we must also subtract the force from any operations that are now too far behind to have an effect.
After repeating this process for every single number, there's one last check.
If our 'reduction force' is exactly zero after we've passed the final number, it means all operations concluded neatly and the task is possible. Otherwise, it's not.

Code Implementation

def make_array_elements_equal_to_zero(numbers_list, operation_length_k):
    # This running tally tracks the sum of all subtraction waves currently affecting an element.
    current_subtraction_tally = 0

    for current_index in range(len(numbers_list)):
        # A subtraction wave's effect is finite; we undo its contribution once we are past its end point.
        if current_index >= operation_length_k:
            current_subtraction_tally -= numbers_list[current_index - operation_length_k]

        effective_value = numbers_list[current_index] - current_subtraction_tally

        # An element's value must not become negative, as we can only ever subtract, not add.
        if effective_value < 0:
            return False

        # This effective value is the new subtraction wave we must start, so we add it to the tally.
        current_subtraction_tally += effective_value
        numbers_list[current_index] = effective_value

    # Any remaining tally implies operations would affect elements beyond the array, which is invalid.
    return current_subtraction_tally == 0

Big(O) Analysis

Time Complexity

O(n) – The algorithm's runtime is governed by a single iteration through the input array of size n. For each of the n elements, the solution performs a small, constant number of arithmetic operations and comparisons. This involves updating the element based on a running 'reduction force', and then adjusting that force based on the element's new value and the value of an element k positions prior. Because every element is visited exactly once, the total work is proportional to n, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm operates in a single pass using a constant number of variables, primarily one to track the current 'reduction force'. The plain English step of needing to subtract the force of operations that are now 'too far behind' can be implemented without an auxiliary data structure. This is typically achieved by reusing the input array itself to store the magnitude of operations started at each index. Since the solution only requires a few variables whose memory footprint is independent of the input size N, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Input array with a single element, which must be zero e.g., [0]	The simulation correctly handles moving out of bounds immediately, resulting in a valid case as the array is already all zeros.
Input array where all elements are already zero	The simulation simply moves the cursor in the chosen direction until it goes out of bounds, correctly identifying any start as valid.
A single large non-zero element adjacent to the only starting zero, e.g., [100, 0]	The simulation correctly models the oscillation required to decrement the large number, distinguishing the valid starting direction from the invalid one.
Symmetrical non-zero elements around a central zero, e.g., [1, 0, 1]	A correct simulation will track the cursor as it crosses the starting zero multiple times to clear elements on both sides.
An unbalanced configuration where no solution exists, e.g., [2, 0, 1]	The simulation will run for all potential starts, but in each case, it will terminate with non-zero elements remaining in the array.
Multiple, separated zeros providing different starting points	The solution must iterate through every possible starting zero and direction, as the outcome can be different for each choice.
Inputs at maximum constraints (N=100, nums[i]=100)	The direct simulation approach must be efficient enough to handle the maximum number of steps, which can be up to O(sum(nums) * N).
A starting zero is at an edge of the array, e.g., [0, 5, 5]	Choosing to move away from the other elements will cause immediate termination, which the simulation correctly identifies as an invalid choice.

Input array with a single element, which must be zero e.g., [0]

How to Handle:

The simulation correctly handles moving out of bounds immediately, resulting in a valid case as the array is already all zeros.

Input array where all elements are already zero

How to Handle:

The simulation simply moves the cursor in the chosen direction until it goes out of bounds, correctly identifying any start as valid.

A single large non-zero element adjacent to the only starting zero, e.g., [100, 0]

How to Handle:

The simulation correctly models the oscillation required to decrement the large number, distinguishing the valid starting direction from the invalid one.

Symmetrical non-zero elements around a central zero, e.g., [1, 0, 1]

How to Handle:

A correct simulation will track the cursor as it crosses the starting zero multiple times to clear elements on both sides.

An unbalanced configuration where no solution exists, e.g., [2, 0, 1]

How to Handle:

The simulation will run for all potential starts, but in each case, it will terminate with non-zero elements remaining in the array.

Multiple, separated zeros providing different starting points

How to Handle:

The solution must iterate through every possible starting zero and direction, as the outcome can be different for each choice.

Inputs at maximum constraints (N=100, nums[i]=100)

How to Handle:

The direct simulation approach must be efficient enough to handle the maximum number of steps, which can be up to O(sum(nums) * N).

A starting zero is at an edge of the array, e.g., [0, 5, 5]

How to Handle:

Choosing to move away from the other elements will cause immediate termination, which the simulation correctly identifies as an invalid choice.