900. RLE Iterator LeetCode Solution | Easy Approach

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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 iencoding[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.

RLE Iterator Solutions

Time: O(n)
Space: O(n)

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