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This approach involves scanning through the array once to check for any zeros and counting the number of negative numbers. If there's a zero, the product is zero. Otherwise, if the number of negative numbers is odd, the product is negative; if even, the product is positive.
Time Complexity: O(n), where n is the number of elements in the array. We only need to pass over the array once.
Space Complexity: O(1), as no additional space beyond the given variables is used.
1using System;
2
3public class Solution {
4 public static int ArraySign(int[] nums) {
5 int negativeCount = 0;
6 foreach (int num in nums) {
7 if (num == 0) return 0;
8 if (num < 0) negativeCount++;
9 }
10 return (negativeCount % 2 == 0) ? 1 : -1;
11 }
12
13 public static void Main() {
14 int[] nums = {-1, -2, -3, -4, 3, 2, 1};
15 Console.WriteLine(ArraySign(nums));
16 }
17}The C# solution follows the general pattern with C# idioms, using a foreach loop to count negative and zero elements.
This approach uses a multiplicative identity set as variable 'sign', initialized to 1. Then iterates through each element of the array, multiplying 'sign' with -1 for each negative number and returning zero if a zero is found.
Time Complexity: O(n)
Space Complexity: O(1)
1
Python uses a similar pattern, where an initialized 'sign' is multiplied by -1 for every negative, zero results in an immediate 0 return.