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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.

  • 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); */ 
  • 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)  
  • 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) */ 

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