public class RLEIterator extends Object
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.
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”]
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 <= 1000encoding.length is even.0 <= encoding[i] <= 1091 <= n <= 1091000 calls will be made to next.| Constructor and Description |
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RLEIterator(int[] encoding) |
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