Find the Index of the Large Integer

lcpy gen -n 1533   # by problem number
lcpy gen -s find_the_index_of_the_large_integer   # by problem name

Input: arr = [7,7,7,7,10,7,7,7]
Output: 4
Explanation: 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.

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

class ArrayReader:
    # Test-harness API: backs the compareSub/length interface with the array
    def __init__(self, arr: list[int]) -> None:
        self.arr = arr

    def compare_sub(self, lo1: int, hi1: int, lo2: int, hi2: int) -> int:
        left = sum(self.arr[lo1 : hi1 + 1])
        right = sum(self.arr[lo2 : hi2 + 1])
        return (left > right) - (left < right)

    def length(self) -> int:
        return len(self.arr)


class Solution:
    # Time: O(log n) compare_sub calls
    # Space: O(1)
    def get_index(self, reader: ArrayReader) -> int:
        left, right = 0, reader.length() - 1
        while left < right:
            # Split into two equal-size leading blocks plus a remainder block;
            # equal sizes guarantee a 0 result rules out both leading blocks.
            t2 = left + (right - left) // 3
            t3 = left + ((right - left) // 3) * 2 + 1
            cmp = reader.compare_sub(left, t2, t2 + 1, t3)
            if cmp == 0:
                left = t3 + 1
            elif cmp == 1:
                right = t2
            else:
                left, right = t2 + 1, t3
        return left