Sponsored
Sponsored
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.
1def arraySign(nums):
2 negative_count = 0
3 for num in nums:
4 if num == 0:
5 return 0
6 if num < 0:
7 negative_count += 1
8 return 1 if negative_count % 2 == 0 else -1
9
10print(arraySign([-1, -2, -3, -4, 3, 2, 1]))The Python solution maintains a similar process to other languages with Pythonic constructs. The list is iterated, and the return depends on zero presence and negative count parity.
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
public class Solution {
public static int ArraySign(int[] nums) {
int sign = 1;
foreach (int num in nums) {
if (num == 0) return 0;
if (num < 0) sign *= -1;
}
return sign;
}
public static void Main() {
int[] nums = {-1, -2, -3, -4, 3, 2, 1};
Console.WriteLine(ArraySign(nums));
}
}This C# code utilizes a sign flipping logic guided by loop-over methods to incrementally adjust result sign.