900. RLE Iterator

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.

  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] <= 10^9

• 1 <= n <= 10^9

• At most 1000 calls will be made to next.

Solution (Java)

class RLEIterator {
    private int index;
    private final int[] array;

    public RLEIterator(int[] encoding) {
        index = 0;
        array = encoding;
    }

    public int next(int n) {
        int lastElement = -1;
        while (n > 0 && index < array.length) {
            if (array[index] > n) {
                array[index] -= n;
                lastElement = array[index + 1];
                break;
            } else if (array[index] == n) {
                array[index] = 0;
                lastElement = array[index + 1];
                index += 2;
                break;
            } else {
                n -= array[index];
                index += 2;
            }
        }
        return lastElement;
    }
}

/**
 * Your RLEIterator object will be instantiated and called as such:
 * RLEIterator obj = new RLEIterator(encoding);
 * int param_1 = obj.next(n);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).