README_EN.md

900. RLE Iterator

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Description

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.

Solutions

Python3

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)

Java

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);
 */

C++

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);
 */

Go

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);
 */

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