In a given integer array `nums`

, there is always exactly one largest element.

Find whether the largest element in the array is at least twice as much as every other number in the array.

If it is, return the **index** of the largest element, otherwise return -1.

**Example 1**:

Input: nums = [3, 6, 1, 0]

Output: 1

Explanation: 6 is the largest integer, and for every other number in the array x, 6 is more than twice as big as x. The index of value 6 is 1, so we return 1.

**Example 2**:

Input: nums = [1, 2, 3, 4]

Output: -1

Explanation: 4 isn’t at least as big as twice the value of 3, so we return -1.

**Note**:

`nums`

will have a length in the range`[1, 50]`

.- Every
`nums[i]`

will be an integer in the range`[0, 99]`

.

## Solution

```
from typing import List
class Solution:
def dominantIndex(self, nums: List[int]) -> int:
largestNumIndex = -1
largestNum = 0
secondLargestNum = 0
for index, num in enumerate(nums):
if num >= largestNum:
secondLargestNum = largestNum
largestNum = num
largestNumIndex = index
elif num > secondLargestNum:
secondLargestNum = num
if largestNum >= secondLargestNum * 2:
return largestNumIndex
return -1
```

## Test Cases

```
test = Solution()
answer = test.dominantIndex([1,2,3,4])
assert answer == -1
answer = test.dominantIndex([1])
assert answer == 0
answer = test.dominantIndex([1,2])
assert answer == 1
answer = test.dominantIndex([0,99])
assert answer == 1
answer = test.dominantIndex([0,0,3,2])
assert answer == -1
answer = test.dominantIndex([49,99])
assert answer == 1
print('All Passed!')
```

## Big O Analysis

**Space Complexity**: O(1)

**Time Complexity**: O(N)