**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 :

` ````
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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