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We have an integer array `arr`

, where all the integers in `arr`

are equal except for one integer which is **larger** than the rest of the integers. You will not be given direct access to the array, instead, you will have an **API** `ArrayReader`

which have the following functions:

`int compareSub(int l, int r, int x, int y)`

: where`0 <= l, r, x, y < ArrayReader.length()`

,`l <= r and`

`x <= y`

. The function compares the sum of sub-array`arr[l..r]`

with the sum of the sub-array`arr[x..y]`

and returns:**1**if`arr[l]+arr[l+1]+...+arr[r] > arr[x]+arr[x+1]+...+arr[y]`

.**0**if`arr[l]+arr[l+1]+...+arr[r] == arr[x]+arr[x+1]+...+arr[y]`

.**-1**if`arr[l]+arr[l+1]+...+arr[r] < arr[x]+arr[x+1]+...+arr[y]`

.

`int length()`

: Returns the size of the array.

You are allowed to call `compareSub()`

**20 times** at most. You can assume both functions work in `O(1)`

time.

Return *the index of the array arr which has the largest integer*.

**Follow-up:**

- What if there are two numbers in
`arr`

that are bigger than all other numbers? - What if there is one number that is bigger than other numbers and one number that is smaller than other numbers?

**Example 1:**

Input:arr = [7,7,7,7,10,7,7,7]Output:4Explanation:The following calls to the API reader.compareSub(0, 0, 1, 1) // returns 0 this is a query comparing the sub-array (0, 0) with the sub array (1, 1), (i.e. compares arr[0] with arr[1]). Thus we know that arr[0] and arr[1] doesn't contain the largest element. reader.compareSub(2, 2, 3, 3) // returns 0, we can exclude arr[2] and arr[3]. reader.compareSub(4, 4, 5, 5) // returns 1, thus for sure arr[4] is the largest element in the array. Notice that we made only 3 calls, so the answer is valid.

**Example 2:**

Input:nums = [6,6,12]Output:2

**Constraints:**

`2 <= arr.length <= 5 * 10^5`

`1 <= arr[i] <= 100`

- All elements of
`arr`

are equal except for one element which is larger than all other elements.