How do you solve Make Array Elements Equal to Zero on LeetCode?

Make Array Elements Equal to Zero (LeetCode #3354) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The number of zeros in the array directly influences the number of valid starting positions.

What is the time complexity of Make Array Elements Equal to Zero?

The optimal solution for Make Array Elements Equal to Zero runs in O(n) time and uses O(n) space. This complexity is O(n) because we iterate through the array once and perform a constant amount of work for each zero found.

What is the brute force solution for Make Array Elements Equal to Zero?

The brute force approach for Make Array Elements Equal to Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves simulating the entire process for each valid starting position and direction. We check if all elements can be reduced to zero by following the movement rules. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Make Array Elements Equal to Zero?

Make Array Elements Equal to Zero uses the following concepts: Array, Simulation, Prefix Sum. The recommended patterns to study are: Simulation, Array.

What are the interview tips for Make Array Elements Equal to Zero?

Always clarify the problem statement and constraints before jumping into coding. Think through the simulation process step-by-step to avoid logical errors. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Make Array Elements Equal to Zero?

Failing to account for edge cases where the array has no zeros. Not properly simulating the movement direction changes.

#3354

Make Array Elements Equal to Zero

Easy

ArraySimulationPrefix SumSimulationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach leverages the fact that we can determine valid starting positions based on the number of zeros in the array. We can count how many zeros are present and check if they can cover all non-zero elements.

⚙️

Algorithm

3 steps

1Step 1: Count the number of zeros in the array.
2Step 2: For each zero, check if it can reach all non-zero elements by simulating the process.
3Step 3: Return the count of valid starting positions.

solution.py21 lines

1def countValidSelections(nums):
2    n = len(nums)
3    count = 0
4    for i in range(n):
5        if nums[i] == 0:
6            left, right = i, i
7            temp = nums[:]
8            while left >= 0 or right < n:
9                if left >= 0:
10                    if temp[left] > 0:
11                        temp[left] -= 1
12                        right += 1
13                    left -= 1
14                if right < n:
15                    if temp[right] > 0:
16                        temp[right] -= 1
17                        left -= 1
18                    right += 1
19            if all(x == 0 for x in temp):
20                count += 1
21    return count

ℹ

Complexity note: This complexity is O(n) because we iterate through the array once and perform a constant amount of work for each zero found.

1The number of zeros in the array directly influences the number of valid starting positions.
2Simulating the process helps to understand the movement and how it affects the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.