LargestNumberAtLeastTwiceOfOthers.js

// Source : https://leetcode.com/problems/largest-number-at-least-twice-of-others
// Author : Dean Shi
// Date   : 2018-01-14

/***************************************************************************************
 *
 * 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].
 *
 ***************************************************************************************/

/**
 * @param {number[]} nums
 * @return {number}
 */
var dominantIndex = function(nums) {
    if (nums.length === 1) return 0

    const [largestNum, secondLargestNum] = nums.slice(0).sort((a, b) => b - a)
    return largestNum >= (secondLargestNum * 2) ? nums.indexOf(largestNum) : -1
};