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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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