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Problem solution · Python

Iterator for Combination

Iterator for Combination: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
94 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Iterator for Combination, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 94 lines of Python from the credited upstream file iterator-for-combination.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeIterator for Combination · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(k), per operation# Space: O(k) import itertools  class CombinationIterator(object):     def __init__(self, characters, combinationLength):        """        :type characters: str        :type combinationLength: int        """        self.__it = itertools.combinations(characters, combinationLength)        self.__curr = None        self.__last = characters[-combinationLength:]     def next(self):        """        :rtype: str        """        self.__curr = "".join(self.__it.next())        return self.__curr        def hasNext(self):        """        :rtype: bool        """        return self.__curr != self.__last  # Time:  O(k), per operation# Space: O(k)import functools  class CombinationIterator2(object):     def __init__(self, characters, combinationLength):        """        :type characters: str        :type combinationLength: int        """        self.__characters = characters        self.__combinationLength = combinationLength        self.__it = self.__iterative_backtracking()        self.__curr = None        self.__last = characters[-combinationLength:]            def __iterative_backtracking(self):        def conquer():            if len(curr) == self.__combinationLength:                return curr         def prev_divide(c):            curr.append(c)                def divide(i):            if len(curr) != self.__combinationLength:                for j in reversed(xrange(i, len(self.__characters)-(self.__combinationLength-len(curr)-1))):                    stk.append(functools.partial(post_divide))                    stk.append(functools.partial(divide, j+1))                    stk.append(functools.partial(prev_divide, self.__characters[j]))            stk.append(functools.partial(conquer))         def post_divide():            curr.pop()                    curr = []        stk = [functools.partial(divide, 0)]        while stk:            result = stk.pop()()            if result is not None:                yield result     def next(self):        """        :rtype: str        """        self.__curr = "".join(next(self.__it))        return self.__curr            def hasNext(self):        """        :rtype: bool        """        return self.__curr != self.__last  # Your CombinationIterator object will be instantiated and called as such:# obj = CombinationIterator(characters, combinationLength)# param_1 = obj.next()# param_2 = obj.hasNext()

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