LeetCode 0900. RLE Iterator Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++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)(code-box)

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); */(code-box)

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); */(code-box)

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); */(code-box)

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) */(code-box)