Formatted question description: https://leetcode.ca/all/900.html

900. RLE Iterator (Medium)

Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence. More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.

The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way. If there is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5]. This is because the sequence can be read as "three eights, zero nines, two fives".

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 is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:

.next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].

.next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].

.next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].

.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.

Note:

0 <= A.length <= 1000
A.length is an even integer.
0 <= A[i] <= 10^9
There are at most 1000 calls to RLEIterator.next(int n) per test case.
Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

Related Topics:
Array

Solution 1.

class RLEIterator {
    private int[] encoding;
    private int curr;
    private int i;

    public RLEIterator(int[] encoding) {
        this.encoding = encoding;
        curr = 0;
        i = 0;
    }

    public int next(int n) {
        while (i < encoding.length) {
            if (curr + n > encoding[i]) {
                n -= encoding[i] - curr;
                i += 2;
                curr = 0;
            } else {
                curr += n;
                return encoding[i + 1];
            }
        }
        return -1;
    }
}

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

class RLEIterator {
public:
    vector<int> encoding;
    int curr;
    int i;

    RLEIterator(vector<int>& encoding) {
        this->encoding = encoding;
        this->curr = 0;
        this->i = 0;
    }

    int next(int n) {
        while (i < encoding.size()) {
            if (curr + n > encoding[i]) {
                n -= encoding[i] - curr;
                curr = 0;
                i += 2;
            } else {
                curr += n;
                return encoding[i + 1];
            }
        }
        return -1;
    }
};

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

class RLEIterator:
    def __init__(self, encoding: List[int]):
        self.encoding = encoding
        self.i = 0
        self.curr = 0

    def next(self, n: int) -> int:
        while self.i < len(self.encoding):
            if self.curr + n > self.encoding[self.i]:
                n -= self.encoding[self.i] - self.curr
                self.curr = 0
                self.i += 2
            else:
                self.curr += n
                return self.encoding[self.i + 1]
        return -1


# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(encoding)
# param_1 = obj.next(n)

type RLEIterator struct {
	encoding []int
	curr     int
	i        int
}

func Constructor(encoding []int) RLEIterator {
	return RLEIterator{encoding: encoding, curr: 0, i: 0}
}

func (this *RLEIterator) Next(n int) int {
	for this.i < len(this.encoding) {
		if this.curr+n > this.encoding[this.i] {
			n -= this.encoding[this.i] - this.curr
			this.curr = 0
			this.i += 2
		} else {
			this.curr += n
			return this.encoding[this.i+1]
		}
	}
	return -1
}

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

900 - RLE Iterator

900. RLE Iterator (Medium)

Solution 1.

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900. RLE Iterator (Medium)

Solution 1.

All Problems

All Solutions