0900.RLE_Iterator.py

"""
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.
"""
class RLEIterator:

    def __init__(self, encoding: List[int]):
        self.st = []
        for i in range(len(encoding) - 2, -1, -2):
            if encoding[i] == 0:
                continue
            self.st.append((encoding[i+1], encoding[i])) #store ch, fre

    def next(self, n: int) -> int:
        while n > 0:
            if len(self.st) == 0:
                return -1
            if self.st[-1][1] < n:
                num, fre = self.st.pop()
                n -= fre
            elif self.st[-1][1] == n:
                num, fre = self.st.pop()
                return num
            else:
                num, fre = self.st.pop()
                self.st.append((num, fre - n))
                return self.st[-1][0]


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