Question:Given an array, there is a largest element N. Check if that number is at least twice than all the other elements in the array. Return the index if it is, else return -1Input:{3, 6, 1, 0}Output:-1

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.

Let us try to make one more example**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.

**Algorithm:**

– Scan through the array to find the unique largest element ‘N’, keeping track of it’s index maxIndex.

– Scan through the array again. If we find some x != N with N < 2 * x, we should return -1.

– Otherwise, we should return maxIndex.

```
public int twiceIndex(int[] arr) {
int maxIndex = 0;
for (int i = 0; i < arr.length; ++i) {
if (arr[i] > arr[maxIndex])
maxIndex = i;
}
for (int i = 0; i < arr.length; ++i) {
if (maxIndex != i && arr[maxIndex] < 2 * arr[i])
return -1;
}
return maxIndex;
}
```

*Time Complexity:* O(N) where N is the length of array.*Space Complexity:* O(1), the space used by our int variables.

A working solution can be found here:- https://ideone.com/uyEpv9