README_EN.md

747. Largest Number At Least Twice of Others

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Description

You are given an integer array nums where the largest integer is unique.

Determine 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, or return -1 otherwise.

 

Example 1:

Input: nums = [3,6,1,0]
Output: 1
Explanation: 6 is the largest integer.
For every other number in the array x, 6 is at least 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 is less than twice the value of 3, so we return -1.

Example 3:

Input: nums = [1]
Output: 0
Explanation: 1 is trivially at least twice the value as any other number because there are no other numbers.

 

Constraints:

• 1 <= nums.length <= 50
• 0 <= nums[i] <= 100
• The largest element in nums is unique.

Solutions

Python3

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

Java

class Solution {
    public int dominantIndex(int[] nums) {
        int mx = 0, mid = 0;
        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;
    }
}

C++

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        int mx = 0, mid = 0;
        int ans = 0;
        for (int i = 0; i < nums.size(); ++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;
    }
};

Go

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
}

JavaScript

/**
 * @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;
};

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