Question

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

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.

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)
 */