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 <= 1000encoding.length is even.0 <= encoding[i] <= 1091 <= 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.
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| 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)
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| 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);
*/
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| 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);
*/
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| 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);
*/
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| 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)
*/
|
