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Sign of the Product of an Array

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).

 

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

Solution Explanation: Sign of the Product of an Array

The problem asks to determine the sign of the product of all numbers in a given integer array. The sign can be positive (1), negative (-1), or zero (0).

Approach

The solution employs a straightforward iterative approach:

  1. Initialization: A variable ans is initialized to 1 (representing a positive sign).

  2. Iteration: The code iterates through each number (v) in the input array nums.

  3. Zero Check: If a number is 0, the product becomes 0, so the function immediately returns 0.

  4. Negative Check: If a number is negative, the sign of the product flips. Therefore, ans is multiplied by -1.

  5. Result: After iterating through all numbers, ans holds the final sign of the product. The function returns ans.

Time and Space Complexity

  • Time Complexity: O(n), where n is the length of the input array. The algorithm iterates through the array once.
  • Space Complexity: O(1). The algorithm uses a constant amount of extra space, regardless of the input size.

Code Implementation in Multiple Languages

The following code snippets demonstrate the solution in several programming languages:

Python:

class Solution:
    def arraySign(self, nums: List[int]) -> int:
        ans = 1
        for v in nums:
            if v == 0:
                return 0
            if v < 0:
                ans *= -1
        return ans

Java:

class Solution {
    public int arraySign(int[] nums) {
        int ans = 1;
        for (int v : nums) {
            if (v == 0) {
                return 0;
            }
            if (v < 0) {
                ans *= -1;
            }
        }
        return ans;
    }
}

C++:

class Solution {
public:
    int arraySign(vector<int>& nums) {
        int ans = 1;
        for (int v : nums) {
            if (v == 0) return 0;
            if (v < 0) ans *= -1;
        }
        return ans;
    }
};

JavaScript:

/**
 * @param {number[]} nums
 * @return {number}
 */
var arraySign = function(nums) {
    let ans = 1;
    for (let v of nums) {
        if (v === 0) return 0;
        if (v < 0) ans *= -1;
    }
    return ans;
};

Go:

func arraySign(nums []int) int {
    ans := 1
    for _, v := range nums {
        if v == 0 {
            return 0
        }
        if v < 0 {
            ans *= -1
        }
    }
    return ans
}

C#:

public class Solution {
    public int ArraySign(int[] nums) {
        int ans = 1;
        foreach (int v in nums) {
            if (v == 0) return 0;
            if (v < 0) ans *= -1;
        }
        return ans;
    }
}

These examples all follow the same basic logic, adapting the syntax to the specific language. They all achieve a linear time complexity and constant space complexity.