> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# RLE Iterator Python Solution with Tests

> Tested Python solution for LeetCode 900 with 15 pytest cases. Generate a practice environment with lcpy.

LeetCode 900, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [Design](/catalog/topics/design), [Counting](/catalog/topics/counting), Iterator. [View on LeetCode](https://leetcode.com/problems/rle-iterator/description/).

Generate this problem as a practice environment: tested reference solution, 15 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 900   # by problem number
lcpy gen -s rle_iterator   # by problem name
```

## Problem

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.

### Examples

```
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] <= 10^9`
* `1 <= n <= 10^9`
* At most `1000` calls will be made to `next`.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/rle_iterator/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/rle_iterator/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
from bisect import bisect_left


class RLEIterator:
    # Time: O(1) init, O(log k) next where k is the number of runs
    # Space: O(k) for the prefix counts

    def __init__(self, encoding: list[int]) -> None:
        self.prefix: list[int] = []
        self.values = encoding[1::2]
        self.pos: int = 0
        total = 0
        for count in encoding[::2]:
            total += count
            self.prefix.append(total)

    def next(self, n: int) -> int:
        self.pos += n
        idx = bisect_left(self.prefix, self.pos)
        if idx == len(self.prefix):
            return -1
        return self.values[idx]
```

## Complexity

| Time | Space |
| - | - |
| O(1) init, O(log k) next where k is the number of runs | O(k) for the prefix counts |

## Tags


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