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

Solution Explanation

This problem requires designing a CombinationIterator class that generates combinations of a given string. The solutions presented use two different approaches:

Solution 1: Backtracking

This approach uses a recursive backtracking algorithm to generate all possible combinations of length combinationLength from the input string characters. It then stores these combinations in a list and iterates through them using an index.

• __init__(self, characters: str, combinationLength: int): The constructor initializes the iterator. It uses a depth-first search (DFS) function (dfs) to generate all combinations. dfs recursively adds characters to a temporary list (t), adding the complete combination to cs when the desired length is reached.

• next(self) -> str: This method returns the next combination from the pre-generated list cs. It increments the index idx after each call.

• hasNext(self) -> bool: This method checks if there are more combinations to return. It returns True if idx is less than the length of cs, and False otherwise.

Solution 2: Bit Manipulation

This solution leverages bit manipulation for a more concise and potentially efficient implementation.

• __init__(self, characters: str, combinationLength: int): The constructor initializes the iterator. curr is initialized to (1 << len(characters)) - 1, representing the binary representation with all bits set (representing all characters selected). size stores the desired combination length. cs stores the reversed input string for easier bit manipulation.

• next(self) -> str: This method iterates backward from curr, finding the next number with exactly size bits set (representing the next combination). It constructs the combination string from the bits of curr, reverses the string (as characters were stored in reverse order), and decrements curr.

• hasNext(self) -> bool: This method similarly searches for the next valid curr and returns True if it exists, indicating more combinations are available, and False otherwise.

Time and Space Complexity Analysis

Solution 1 (Backtracking):

• Time Complexity: The time complexity is O(N * 2N), where N is the length of the input string characters. This is because the backtracking algorithm explores all possible combinations. The generation of combinations takes exponential time.

• Space Complexity: The space complexity is O(N * 2N) in the worst case to store all generated combinations in the cs list.

Solution 2 (Bit Manipulation):

• Time Complexity: The time complexity is O(N * C(N, K)), where N is the length of the input string and K is combinationLength. The next() and hasNext() methods take time proportional to the number of combinations, which is given by C(N, K), the number of combinations of choosing K items out of N. The construction of a string takes O(K). Overall O(N*C(N,K)).

• Space Complexity: The space complexity is O(N) to store the reversed input string. No extra space is used to generate combinations because it relies on bit manipulation.

Which solution is better?

Solution 2 (Bit manipulation) is generally preferred because it avoids explicitly generating and storing all combinations upfront, making it more memory-efficient for larger input strings. Solution 1 becomes very memory intensive as N grows because it generates all possible combinations first, storing them in memory. Solution 2 is generally faster as it avoids the overhead of recursion. However, the time complexity of both is impacted by the exponential nature of combinations. If only a few next() calls are expected, the backtracking approach might be a simpler implementation.