How do you solve Sign of the Product of an Array on LeetCode?

Sign of the Product of an Array (LeetCode #1822) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The presence of zero in the array immediately determines the product's sign.

What is the brute force solution for Sign of the Product of an Array?

The brute force approach for Sign of the Product of an Array is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves calculating the product of all elements in the array directly. This is straightforward but can lead to overflow issues with large products. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sign of the Product of an Array?

Sign of the Product of an Array uses the following concepts: Array, Math. The recommended patterns to study are: Array, Mathematics.

What are the interview tips for Sign of the Product of an Array?

Always check for edge cases like zeros in the array. Explain your thought process clearly while coding. Consider how you can optimize your solution before jumping into implementation.

What are common mistakes when solving Sign of the Product of an Array?

Forgetting to handle the case where zero is present in the array. Not realizing that the product's sign can be determined without calculating the actual product.

#1822

Sign of the Product of an Array

Easy

ArrayMathArrayMathematics

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of calculating the product directly, we can determine the sign of the product by counting the number of negative numbers and checking for zeros. This avoids overflow and unnecessary calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for negative numbers and a flag for zero.
2Step 2: Iterate through the array and update the counter for negatives and set the zero flag if a zero is found.
3Step 3: If zero is found, return 0. Otherwise, return 1 if the count of negatives is even, otherwise return -1.

solution.py8 lines

1def arraySign(nums):
2    negative_count = 0
3    for num in nums:
4        if num == 0:
5            return 0
6        if num < 0:
7            negative_count += 1
8    return 1 if negative_count % 2 == 0 else -1

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the array once. The space complexity is O(1) since we only use a few extra variables.

1The presence of zero in the array immediately determines the product's sign.
2The sign of the product can be determined by counting negative numbers.

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