900. RLE Iterator

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

Solution 1: Maintain Two Pointers

We define two pointers \(i\) and \(j\), where pointer \(i\) points to the current run-length encoding being read, and pointer \(j\) points to which character in the current run-length encoding is being read. Initially, \(i = 0\), \(j = 0\).

Each time we call next(n), we judge whether the remaining number of characters in the current run-length encoding \(encoding[i] - j\) is less than \(n\). If it is, we subtract \(n\) by \(encoding[i] - j\), add \(2\) to \(i\), and set \(j\) to \(0\), then continue to judge the next run-length encoding. If it is not, we add \(n\) to \(j\) and return \(encoding[i + 1]\).

If \(i\) exceeds the length of the run-length encoding and there is still no return value, it means that there are no remaining elements to be exhausted, and we return \(-1\).

The time complexity is \(O(n + q)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the run-length encoding, and \(q\) is the number of times next(n) is called.

Python3JavaC++GoTypeScript

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

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

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

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

    public int next(int n) {
        while (i < encoding.length) {
            if (encoding[i] - j < n) {
                n -= (encoding[i] - j);
                i += 2;
                j = 0;
            } else {
                j += 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:
    RLEIterator(vector<int>& encoding) {
        this->encoding = encoding;
    }

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

private:
    vector<int> encoding;
    int i = 0;
    int j = 0;
};

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

type RLEIterator struct {
    encoding []int
    i, j     int
}

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

func (this *RLEIterator) Next(n int) int {
    for this.i < len(this.encoding) {
        if this.encoding[this.i]-this.j < n {
            n -= (this.encoding[this.i] - this.j)
            this.i += 2
            this.j = 0
        } else {
            this.j += 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);
 */

class RLEIterator {
    private encoding: number[];
    private i: number;
    private j: number;

    constructor(encoding: number[]) {
        this.encoding = encoding;
        this.i = 0;
        this.j = 0;
    }

    next(n: number): number {
        while (this.i < this.encoding.length) {
            if (this.encoding[this.i] - this.j < n) {
                n -= this.encoding[this.i] - this.j;
                this.i += 2;
                this.j = 0;
            } else {
                this.j += n;
                return this.encoding[this.i + 1];
            }
        }
        return -1;
    }
}

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

900. RLE Iterator

Description

Solutions

Solution 1: Maintain Two Pointers

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