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