Lexicographic Combination Iterator - Amazon Top Interview Questions


Problem Statement :


Implement an iterator where, given a sorted lowercase alphabet string s of unique characters and an integer k:

next() polls the next lexicographically smallest subsequence of length k
hasnext() which returns whether the next subsequence exists

Constraints


n ≤ 100,000 where n is the number of calls to next and hasnext

Example 1

Input

methods = ["constructor", "next", "next", "next", "hasnext"]
arguments = [["abc", 2], [], [], [], []]`

Output

[None, "ab", "ac", "bc", False]


Solution :



title-img 




                        Solution in C++ :

class LexicographicCombinationIterator {
    public:
    LexicographicCombinationIterator(string s, int k) : _k(k), ss(s) {
    }

    string next() {
        if (states.empty()) {
            states.resize(_k);
            iota(states.begin(), states.end(), 0);
        } else {
            int i = get_index();
            states[i++]++;
            for (; i < _k; i++) states[i] = states[i - 1] + 1;
        }
        string ans;
        for (int i : states) ans.push_back(ss[i]);
        return ans;
    }

    bool hasnext() {
        return states.empty() || get_index() >= 0;
    }

    private:
    int get_index() {
        int N = ss.size(), cur = N - 1, i = _k - 1;
        for (; i >= 0 && states[i] == cur; i--, cur--)
            ;
        return i;
    }
    int _k;
    string ss;
    vector<int> states;
};
                    




                        Solution in Python : 
                            
class LexicographicCombinationIterator:
    def __init__(self, s, k):
        self.pointers = [i for i in range(k)]
        self.s = s
        self.k = k
        self.possible = True

    def next(self):
        res = "".join([self.s[i] for i in self.pointers])
        self.possible = False
        for i in range(len(self.pointers) - 1, -1, -1):
            remaining = len(self.pointers) - 1 - i
            if (
                self.pointers[i] + 1 < len(self.s)
                and len(self.s) - 1 - (self.pointers[i] + 1) >= remaining
            ):
                self.pointers[i] += 1
                for j in range(i + 1, len(self.pointers)):
                    self.pointers[j] = self.pointers[j - 1] + 1
                self.possible = True
                break
        return res

    def hasnext(self):
        return self.possible


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