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1822. Sign of the Product of an Array

Description

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

Solutions

Solution 1: Direct Traversal

The problem requires us to return the sign of the product of the array elements, i.e., return $1$ for positive numbers, $-1$ for negative numbers, and $0$ if it equals $0$.

We can define an answer variable ans, initially set to $1$.

Then we traverse each element $v$ in the array. If $v$ is a negative number, we multiply ans by $-1$. If $v$ is $0$, we return $0$ in advance.

After the traversal is over, we return ans.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

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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

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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;
    }
}

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

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

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impl Solution {
    pub fn array_sign(nums: Vec<i32>) -> i32 {
        let mut ans = 1;
        for &num in nums.iter() {
            if num == 0 {
                return 0;
            }
            if num < 0 {
                ans *= -1;
            }
        }
        ans
    }
}

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

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int arraySign(int* nums, int numsSize) {
    int ans = 1;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == 0) {
            return 0;
        }
        if (nums[i] < 0) {
            ans *= -1;
        }
    }
    return ans;
}

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