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1822. Sign of the Product of an Array

Description

There is 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 $O(1)$.

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