## Description

https://leetcode.com/problems/largest-number-at-least-twice-of-others/

You are given an integer array `nums`

where the largest integer is **unique**.

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:1Explanation: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:-1Explanation:4 is not at least as big as twice the value of 3, so we return -1.

**Constraints:**

`1 <= nums.length <= 50`

`0 <= nums[i] <= 100`

- The largest element in
`nums`

is unique.

## Explanation

We can sort first to find if there is a dominant value.

## Python Solution

```
class Solution:
def dominantIndex(self, nums: List[int]) -> int:
if len(nums) < 2:
return 0
sorted_nums = sorted(nums)
if sorted_nums[-1] >= sorted_nums[-2] * 2:
return nums.index(sorted_nums[-1])
return -1
```

- Time Complexity: O(Nlog(N)).
- Space Complexity: O(N).