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

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)