Problem

Implement a function signFunc(x) that returns:

  • 1 if x is positive.
  • -1 if x is negative.
  • 0 if x is equal to 0.

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

Return signFunc(product).

Examples

Example 1

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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

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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

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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

Solution

Method 1 – Count Negatives and Zeros

Intuition

The sign of the product is 0 if any element is 0, otherwise it is -1 if there are an odd number of negatives, else 1.

Approach

  1. Initialize a variable to count negatives.
  2. If any element is 0, return 0.
  3. If the count of negatives is odd, return -1; else return 1.

Code

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#include <vector>
using namespace std;
class Solution {
public:
    int arraySign(vector<int>& nums) {
        int neg = 0;
        for (int x : nums) {
            if (x == 0) return 0;
            if (x < 0) ++neg;
        }
        return neg % 2 ? -1 : 1;
    }
};
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class Solution {
    public int arraySign(int[] nums) {
        int neg = 0;
        for (int x : nums) {
            if (x == 0) return 0;
            if (x < 0) ++neg;
        }
        return neg % 2 == 1 ? -1 : 1;
    }
}
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class Solution:
    def arraySign(self, nums):
        neg = 0
        for x in nums:
            if x == 0:
                return 0
            if x < 0:
                neg += 1
        return -1 if neg % 2 else 1

Complexity

  • ⏰ Time complexity: O(n) — Single pass through the array.
  • 🧺 Space complexity: O(1) — No extra space used.