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].

Given an array of integers, find the largest one and if it at least twice than the second-largest, return its index, otherwise return -1.

C++ of Finding the Dominant Index of Array

The edge cases need to be dealt when the array is empty or containing only one element. In case of empty array, -1 needs to be returned as there is no dominant number. In case of only 1 element, the answer is 0 as it’s the index of the dominant number.

On other cases, we need to maintain the largest and the second largest number, so we can check if the largest number is at least twice of the second largest number. Of course, we need to remember the index of the largest number.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        if (nums.size() == 0) return -1;
        if (nums.size() == 1) return 0;
        int max1 = nums[0];
        int max2 = INT_MIN;
        int index = 0;
        for (int i = 1; i < nums.size(); ++ i) {
            if (nums[i] > max1) {
                max2 = max1;
                max1 = nums[i];
                index = i;
            } else {
                if (nums[i] > max2) {
                    max2 = nums[i];
                }
            }
        }
        return (max1 &t;= 2 * max2) ? index : -1;
    }
};

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        if (nums.size() == 0) return -1;
        if (nums.size() == 1) return 0;
        int max1 = nums[0];
        int max2 = INT_MIN;
        int index = 0;
        for (int i = 1; i < nums.size(); ++ i) {
            if (nums[i] > max1) {
                max2 = max1;
                max1 = nums[i];
                index = i;
            } else {
                if (nums[i] > max2) {
                    max2 = nums[i];
                }
            }
        }
        return (max1 &t;= 2 * max2) ? index : -1;
    }
};

The above C++ code of finding the dominant index/number of the array runs at O(N) complexity and O(1) space.

–EOF (The Ultimate Computing & Technology Blog) —