m0900.py

"""RLE Iterator
Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length
encoding of some sequence. More specifically, for all even i, A[i] tells us the
number of times that the non-negative integer value A[i+1] is repeated in the
sequence.

The iterator supports one function: next(int n), which exhausts the next n
elements (n >= 1) and returns the last element exhausted in this way. If there
is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of
the sequence [8,8,8,5,5].  This is because the sequence can be read as "three
eights, zero nines, two fives".

Example 1:

* Input = [
    "RLEIterator",
    "next",
    "next",
    "next",
    "next",
],
[
    [[3, 8, 0, 9, 2, 5]],
    [2],
    [1],
    [1],
    [2],
]
* Output: [null,8,8,5,-1]
* Explanation:
  * RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
  * This maps to the sequence [8,8,8,5,5].
  * RLEIterator.next is then called 4 times:
    * .next(2) exhausts 2 terms of the sequence, returning 8. The remaining
      sequence is now [8, 5, 5].
    * .next(1) exhausts 1 term of the sequence, returning 8. The remaining
      sequence is now [5, 5].
    * .next(1) exhausts 1 term of the sequence, returning 5. The remaining
      sequence is now [5].
    * .next(2) exhausts 2 terms, returning -1.  This is because the first term
      exhausted was 5, but the second term did not exist.  Since the last term
      exhausted does not exist, we return -1.

Note:

* 0 <= A.length <= 1000
* A.length is an even integer.
* 0 <= A[i] <= 10^9
* There are at most 1000 calls to RLEIterator.next(int n) per test case.
* Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.
"""