# 747\_Largest Number At Least Twice of Others

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:**

1. `nums`will have a length in the range `[1, 50]`.
2. Every `nums[i]`will be an integer in the range `[0, 99]`.

## Solution:

**Idea**:

* If the largest element is at least twice as much as the second largest element, then it's at least twice as much as all the other elements. 
* Use one linear pass to find the largest and second largest element.

**Time Complexity**: $$O(n)$$

**Space Complexity**: $$O(1)$$

```python
def dominantIndex(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    if len(nums) == 1: return 0

    # m1 is the largest
    # m2 is the second largest
    if nums[0] >= nums[1]:
        m1, m2 = 0, 1
    else:
        m1, m2 = 1, 0

    for i in range(2, len(nums)):
        if nums[i] > nums[m1]:
            m2 = m1
            m1 = i
        elif nums[i] > nums[m2]:
            m2 = i
    return m1 if nums[m1] >= nums[m2]*2 else -1
```


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