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Java-based LeetCode algorithm problem solutions, regularly updated
900\. RLE Iterator
Medium
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`.