# 900. RLE Iterator LeetCode Solution | Easy Approach Share:

RLE Iterator 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]], , , , ]
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 . 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`.

Time: O(n)
Space: O(n)

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