algoadvance

Given an integer array nums, return the sign of the product of all values in the array.

• If the product is positive, return 1.
• If the product is negative, return -1.
• If the product is zero, return 0.

Clarifying Questions

1. Input Size: What is the range of the input size?
   • The provided constraints: ( 1 \leq nums.length \leq 1000 )
2. Element Range: What values can the elements in nums take?
   • The provided constraints: ( -100 \leq nums[i] \leq 100 )
3. Edge Cases: Should we assume that the array can be empty?
   • Given the constraints, no. The array will have at least one element.

Strategy

The main idea is to determine the sign of the product without actually computing the product (since it could cause overflow for large arrays). This can be efficiently achieved with the following steps:

1. Initialize a variable to keep track of the count of negative numbers.
2. Iterate through the array:
   • If an element is zero, immediately return 0 (since the product will be zero).
   • If an element is negative, increment the negative count by 1.
3. After the loop, if the count of negative numbers is odd, the product is negative, so return -1.
4. If the count of negative numbers is even, the product is positive, so return 1.

Code

def arraySign(nums: list[int]) -> int:
    negative_count = 0
    
    for number in nums:
        if number == 0:
            return 0
        elif number < 0:
            negative_count += 1
    
    if negative_count % 2 == 0:
        return 1
    else:
        return -1

Time Complexity

• Time Complexity: (O(n)) where (n) is the number of elements in the array. This is because we are processing each element exactly once.
• Space Complexity: (O(1)) because we are using only a fixed amount of additional space irrespective of the input size.

By following this strategy, we ensure that we can determine the sign of the product efficiently without encountering issues related to product overflow.

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