0900. RLE Iterator

https://leetcode.com/problems/rle-iterator

Description

We can use run-length encoding (i.e., RLE) to encode a sequence of integers. In a run-length encoded array of even length encoding (0-indexed), for all even i, encoding[i] tells us the number of times that the non-negative integer value encoding[i + 1] is repeated in the sequence.

• For example, the sequence arr = [8,8,8,5,5] can be encoded to be encoding = [3,8,2,5]. encoding = [3,8,0,9,2,5] and encoding = [2,8,1,8,2,5] are also valid RLE of arr.

Given a run-length encoded array, design an iterator that iterates through it.

Implement the RLEIterator class:

• RLEIterator(int[] encoded) Initializes the object with the encoded array encoded.

• int next(int n) Exhausts the next n elements and returns the last element exhausted in this way. If there is no element left to exhaust, return -1 instead.

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 rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5].
rLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now [5].
rLEIterator.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.

Constraints:

• 2 <= encoding.length <= 1000

• encoding.length is even.

• 0 <= encoding[i] <= 109

• 1 <= n <= 109

• At most 1000 calls will be made to next.

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