936. Rle Iterator¶

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936. 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.

Solution¶

class RLEIterator {
    private Deque<Pair> dq;
    static class Pair {
        int count, node;
        public Pair(int count, int node) {
            this.count = count;
            this.node = node;
        }
    }
    public RLEIterator(int[] encoding) {
        dq = new ArrayDeque<>();
        int idx = 0, n = encoding.length;
        while (idx < n) {
            dq.addLast(new Pair(encoding[idx], encoding[idx + 1]));
            idx += 2;
        }
    }
    public int next(int n) {
        int req = n;
        int last = -1;
        while (dq.size() > 0 && req > 0) {
            Pair current = dq.pollFirst();
            if (current.count > req) {
                dq.addFirst(new Pair(current.count - req, current.node));
                req = 0;
            }
            else {
                req -= current.count;
            }
            last = current.node;
        }
        if (req > 0) return -1;
        return last;
    }
}

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

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here