Problem URL: Largest number at least twice of others

We will keep track of both largest and second largest number while iterating through the whole nums array. Then if the largest number is at least twice as big as the second largest, then we return the index of largest number, else we return -1.

class Solution:
    def dominantIndex(self, nums: List[int]) -> int:
        largest, second = -1, -1

        for num in nums:
            if num >= largest:
                second = largest
                largest = num
            elif num >= second:
                second = num

        return nums.index(largest) if largest >= second*2 else -1

Time Complexity: O(n)
Space Complexity: O(1)