1286. Iterator for Combination

Problem Link

Description

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Design the CombinationIterator class:

• CombinationIterator(string characters, int combinationLength) Initializes the object with a string characters of sorted distinct lowercase English letters and a number combinationLength as arguments.
• next() Returns the next combination of length combinationLength in lexicographical order.
• hasNext() Returns true if and only if there exists a next combination.

 

Example 1:

Input
["CombinationIterator", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[["abc", 2], [], [], [], [], [], []]
Output
[null, "ab", true, "ac", true, "bc", false]

Explanation
CombinationIterator itr = new CombinationIterator("abc", 2);
itr.next();    // return "ab"
itr.hasNext(); // return True
itr.next();    // return "ac"
itr.hasNext(); // return True
itr.next();    // return "bc"
itr.hasNext(); // return False

 

Constraints:

• 1 <= combinationLength <= characters.length <= 15
• All the characters of characters are unique.
• At most 104 calls will be made to next and hasNext.
• It is guaranteed that all calls of the function next are valid.

Solution

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Python3

class CombinationIterator:
 
    def __init__(self, characters: str, combinationLength: int):
        self.perms = []
        self.n = len(characters)
        self.k = combinationLength
        self.index = 0
        
        def go(i, curr):
            if len(curr) == self.k:
                self.perms.append(curr)
            
            if i == self.n:
                return
 
            for j in range(i, self.n):
                go(j + 1, curr + characters[j])
 
        go(0, '')
 
    def next(self) -> str:
        res = self.perms[self.index]
        self.index += 1
        return res
 
    def hasNext(self) -> bool:
        return self.index < len(self.perms)
 
 
# Your CombinationIterator object will be instantiated and called as such:
# obj = CombinationIterator(characters, combinationLength)
# param_1 = obj.next()
# param_2 = obj.hasNext()