# Leetcode – Sign of the Product of an Array Solution

Leetcode provides a plethora of programming problems designed to help individuals refine their coding skills and prepare for technical interviews. One such problem is called “Sign of the Product of an Array,” which is an interesting exercise for honing Python programming skills, particularly in the use of lists, iteration, and basic mathematical principles. In this extensive article, we will dissect multiple approaches to solving this problem using Python.

## Problem Statement

You are given an integer array nums. Let product be the product of all values in the array. Return the sign of product.

### Constraints

• The array contains up to 1000 elements.
• Each element of the array is an integer between -100 and 100.

### Examples

1. Input: [1, 5, 0, 2, -3] Output: 0
2. Input: [-1, -2, -3, -4, 3, 2, 1] Output: -1
3. Input: [1, 1, 1, 1, 1] Output: 1

## Naive Approach: Simple Iteration and Multiplication

One basic approach to solve this problem is to iterate over the array, multiplying each element to get the total product, and then returning the sign of that product.

def arraySign(nums):
product = 1
for num in nums:
product *= num

if product > 0:
return 1
elif product < 0:
return -1
else:
return 0

### Time Complexity

This naive approach has a time complexity of O(n), where n is the length of the array.

### Space Complexity

The space complexity is O(1), as we’re only using a single variable to store the product.

## Optimized Approach: Early Exit

We don’t necessarily have to calculate the entire product to determine its sign. If we encounter a zero in the array, we can exit early, knowing that the product will be zero.

def arraySign(nums):
for num in nums:
if num == 0:
return 0

product = 1
for num in nums:
product *= num

return 1 if product > 0 else -1

### Time and Space Complexity

This approach still has a time complexity of O(n) but allows for an early exit, making it more efficient in certain scenarios. The space complexity remains O(1).

## Further Optimized Approach: Counting Negatives

We can solve this problem without actually calculating the product by simply counting the number of negative numbers in the array. If the count is odd, the sign of the product will be negative; otherwise, it will be positive.

def arraySign(nums):
count_negative = 0
for num in nums:
if num == 0:
return 0
if num < 0:
count_negative += 1

return -1 if count_negative % 2 != 0 else 1

### Time and Space Complexity

The time complexity is O(n), and the space complexity is O(1).

## Unit Testing

It’s important to test the function against multiple test cases to ensure its correctness.

def test_arraySign():
assert arraySign([1, 5, 0, 2, -3]) == 0
assert arraySign([-1, -2, -3, -4, 3, 2, 1]) == -1
assert arraySign([1, 1, 1, 1, 1]) == 1
print("All test cases pass")

test_arraySign()

## Conclusion

The “Sign of the Product of an Array” problem on Leetcode is an excellent exercise for understanding list iteration, basic mathematical principles, and optimization techniques in Python. Multiple approaches can solve this problem, ranging from naive multiplication to more efficient methods that allow for early exits or avoid multiplication altogether. Each approach has its pros and cons in terms of time and space complexity, providing a valuable lesson in trade-offs and optimization.