# 1538. Guess the Majority in a Hidden Array

https://leetcode.com/problems/guess-the-majority-in-a-hidden-array

## Description

We have an integer array `nums`, where all the integers in `nums` are **0** or **1**. You will not be given direct access to the array, instead, you will have an **API** `ArrayReader` which have the following functions:

* `int query(int a, int b, int c, int d)`: where `0 <= a < b < c < d < ArrayReader.length()`. The function returns the distribution of the value of the 4 elements and returns:
  * **4** : if the values of the 4 elements are the same (0 or 1).
  * **2** : if three elements have a value equal to 0 and one element has value equal to 1 or vice versa.
  * **0**: if two element have a value equal to 0 and two elements have a value equal to 1.
* `int length()`: Returns the size of the array.

You are allowed to call `query()` **2 \* n times** at most where n is equal to `ArrayReader.length()`.

Return **any** index of the most frequent value in `nums`, in case of tie, return -1.

**Follow up:** What is the minimum number of calls needed to find the majority element?

**Example 1:**

```
**Input:** nums = [0,0,1,0,1,1,1,1]
**Output:** 5
**Explanation:** The following calls to the API
reader.length() // returns 8 because there are 8 elements in the hidden array.
reader.query(0,1,2,3) // returns 2 this is a query that compares the elements nums[0], nums[1], nums[2], nums[3]
// Three elements have a value equal to 0 and one element has value equal to 1 or viceversa.
reader.query(4,5,6,7) // returns 4 because nums[4], nums[5], nums[6], nums[7] have the same value.
we can infer that the most frequent value is found in the last 4 elements.
Index 2, 4, 6, 7 is also a correct answer.
```

**Example 2:**

```
**Input:** nums = [0,0,1,1,0]
**Output:** 0
```

**Example 3:**

```
**Input:** nums = [1,0,1,0,1,0,1,0]
**Output:** -1
```

**Constraints:**

* `5 <= nums.length <= 10^5`
* `0 <= nums[i] <= 1`

## ac

```java
```
