Design Compressed String Iterator — LeetCode 604 Python Solution

The problem

Design and implement a data structure for a compressed string iterator. The given compressed string will be in the form of each letter followed by a positive integer representing the number of this letter existing in the original uncompressed string.

Example

Input

["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"]

Output

[null, "L", "e", "e", "t", "C", "o", true, "d", true]

Explanation

StringIterator stringIterator = new StringIterator("L1e2t1C1o1d1e1");

Python solution

class StringIterator:
    def __init__(self, compressedString: str):
        self.d = []
        self.p = 0
        n = len(compressedString)
        i = 0
        while i < n:
            c = compressedString[i]
            x = 0
            i += 1
            while i < n and compressedString[i].isdigit():
                x = x * 10 + int(compressedString[i])
                i += 1
            self.d.append([c, x])

    def next(self) -> str:
        if not self.hasNext():
            return ' '
        ans = self.d[self.p][0]
        self.d[self.p][1] -= 1
        if self.d[self.p][1] == 0:
            self.p += 1
        return ans

    def hasNext(self) -> bool:
        return self.p < len(self.d) and self.d[self.p][1] > 0


# Your StringIterator object will be instantiated and called as such:
# obj = StringIterator(compressedString)
# param_1 = obj.next()
# param_2 = obj.hasNext()

Complexity

Pattern: Hash Map

Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 604. Design Compressed String Iterator is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.

The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.

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