Sign of the Product of an Array

Problem Statement

In this task, you are required to implement a function named signFunc(x) which is responsible for determining the mathematical sign of a given integer x. The function should return:

• 1 if x is positive.
• -1 if x is negative.
• 0 if x is precisely zero.

Additionally, you are provided with an integer array called nums. Your goal is to compute the product of all elements in this array and then pass this product to the signFunc(x) function to get the sign of the product. The result of this operation (the output of signFunc(x) for the computed product) should be returned as the final answer.

Examples

Example 1

Input:

nums = [-1,-2,-3,-4,3,2,1]

Output:

1

Explanation:

The product of all values in the array is 144, and signFunc(144) = 1

Example 2

Input:

nums = [1,5,0,2,-3]

Output:

0

Explanation:

The product of all values in the array is 0, and signFunc(0) = 0

Example 3

Input:

nums = [-1,1,-1,1,-1]

Output:

-1

Explanation:

The product of all values in the array is -1, and signFunc(-1) = -1

Constraints

• 1 <= nums.length <= 1000
• -100 <= nums[i] <= 100

Approach and Intuition

To solve this problem:

1. Understand the behavior of multiplication regarding positive, negative numbers, and zero.

   • Multiplying two negative numbers gives a positive result.
   • Multiplying a negative by a positive number gives a negative result.
   • Any number multiplied by zero results in zero.

2. Based on this understanding, evaluate the sign of the product without necessarily computing the exact product:

   • Iterate through each number in the nums array.
   • Count the number of negative numbers.
   • Check if any number in the array is zero, since that will automatically make the product zero.

3. Determine the result based on counts:

   • If any number in the array is zero, return 0 immediately.
   • If the count of negative numbers is odd, the product will be negative, so return -1.
   • If the count of negative numbers is even, the product will be positive, so return 1.

This method avoids potentially large number multiplication by directly determining the product's sign from the individual elements' properties. This approach ensures that the solution remains efficient and straightforward, even for the upper limit constraints.

Solutions

class Solution {
public:
    int productSign(vector<int>& arr) {
        int resultSign = 1;
        for (int n : arr) {
            if (n == 0) {
                return 0;
            }
            if (n < 0) {
                resultSign *= -1;
            }
        }
    
        return resultSign;
    }
};

class Solution {
    public int productSign(int[] arr) {
        int productIndicator = 1;
        for (int element : arr) {
            if (element == 0) {
                return 0;
            }
            if (element < 0) {
                productIndicator *= -1;
            }
        }
    
        return productIndicator;
    }
}