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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 $O(1)$.

• Java
• C++
• Python
• Go
• Javascript
• RenderScript

• 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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