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Formatted question description: https://leetcode.ca/all/747.html

# 747. Largest Number At Least Twice of Others (Easy)

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 1.

• class Solution {
public int dominantIndex(int[] nums) {
int length = nums.length;
if (length == 1)
return 0;
int[][] numIndexArray = new int[length][2];
for (int i = 0; i < length; i++) {
numIndexArray[i][0] = nums[i];
numIndexArray[i][1] = i;
}
Arrays.sort(numIndexArray, new Comparator<int[]>() {
public int compare(int[] array1, int[] array2) {
if (array1[0] != array2[0])
return array1[0] - array2[0];
else
return array1[1] - array2[1];
}
});
int maxNum = numIndexArray[length - 1][0], maxIndex = numIndexArray[length - 1][1];
int secondMaxNum = numIndexArray[length - 2][0];
return maxNum >= 2 * secondMaxNum ? maxIndex : -1;
}
}

############

class Solution {
public int dominantIndex(int[] nums) {
int mx = Integer.MIN_VALUE, mid = Integer.MIN_VALUE;
int ans = -1;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] > mx) {
mid = mx;
mx = nums[i];
ans = i;
} else if (nums[i] > mid) {
mid = nums[i];
}
}
return mx >= mid * 2 ? ans : -1;
}
}

• // OJ: https://leetcode.com/problems/largest-number-at-least-twice-of-others/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int dominantIndex(vector<int>& nums) {
int first = -1, second = -1;
for (int i = 0; i < nums.size(); ++i) {
if (first == -1 || nums[i] > nums[first]) {
second = first;
first = i;
} else if (second == -1 || nums[i] > nums[second]) second = i;
}
return nums[first] >= nums[second] * 2 ? first : -1;
}
};

• class Solution:
def dominantIndex(self, nums: List[int]) -> int:
mx = mid = 0
ans = -1
for i, v in enumerate(nums):
if v > mx:
mid, mx = mx, v
ans = i
elif v > mid:
mid = v
return ans if mx >= 2 * mid else -1

############

class Solution(object):
def dominantIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 1:
return 0
largest = max(nums)
ind = nums.index(largest)
nums.pop(ind)
if largest >= 2 * max(nums):
return ind
else:
return -1

• func dominantIndex(nums []int) int {
mx, mid := 0, 0
ans := 0
for i, v := range nums {
if v > mx {
mid, mx = mx, v
ans = i
} else if v > mid {
mid = v
}
}
if mx >= mid*2 {
return ans
}
return -1
}

• /**
* @param {number[]} nums
* @return {number}
*/
var dominantIndex = function (nums) {
let mx = 0,
mid = 0;
let ans = 0;
for (let i = 0; i < nums.length; ++i) {
if (nums[i] > mx) {
mid = mx;
mx = nums[i];
ans = i;
} else if (nums[i] > mid) {
mid = nums[i];
}
}
return mx >= mid * 2 ? ans : -1;
};