Problem Description

The CombinationIterator class is designed to handle combinations of a given string. The problem is defined with three requirements:

1. The class must be initialized with a string characters consisting of sorted distinct lowercase English letters and an integer combinationLength which specifies the length of the combinations to be formed.

2. A next() function that, when called, returns the next combination of the specified length combinationLength in lexicographical order.

3. A hasNext() function that returns true if there are more combinations that can be generated and false if no more combinations are available.

The challenges in solving this problem include managing the generation of combinations in the proper order without actual generation of all combinations at once which would be inefficient, as well as determining when no more combinations are available.

Flowchart Walkthrough

First, let's use the algorithm flowchart to figure out the best approach for solving LeetCode problem 1286, Iterator for Combination. Here's a step-by-step analysis following the flowchart:

1. Is it a graph?

   • No - The problem does not deal with nodes and edges typically found in graph structures.

2. Need to solve for kth smallest/largest?

   • No - The task is to create an iterator that returns combinations, not to find a specific kth smallest or largest element.

3. Involves Linked Lists?

   • No - The problem does not involve operations on linked lists.

4. Does the problem have small constraints?

   • Yes - Given that the task is to iterate over combinations, the input constraints (like string length and combination length) allow for some brute force or backtracking methods since all combinations of a relatively small set need to be considered.

5. Brute force / Backtracking?

   • Yes - The problem is inherently about generating combinations from a given set (or string), which is a typical backtracking problem. Each combination needs to be considered, and backtracking allows us to explore all possible combinations of elements dynamically.

Conclusion: Using the flowchart, we deduce that the Backtracking pattern is the appropriate solution for generating combinations iteratively in problem 1286, Iterator for Combination. This is because the problem involves dealing with small constraints and requires generating all possible outcomes (combinations in this case), which is typical for backtracking approaches.

Intuition

To arrive at the solution, the key is to implement an efficient way of generating the next lexicographical combination without having to pre-compute all possible combinations.

The solution code uses bit manipulation to achieve this. Here's how the solution operates:

1. The string characters is reversed and each character is associated with a bit in an integer self.curr. Initially, self.curr is set to 2^len(characters) - 1, which means all the bits are set and then it will decrease with each call to next().

2. The next() function looks for the current combination by checking if there are self.size bits set in self.curr (indicating the current combination's characters). If not, it keeps decrementing self.curr until there are the correct number of bits set, indicating the next valid combination.

3. Once a valid combination is found, the corresponding characters are extracted and joined to form the next lexicographical combination string.

4. The hasNext() function also decrements self.curr while invalid combinations are found (those not having the exact number of desired bits set), checking afterward if there are any valid combinations left (by checking if self.curr is greater than or equal to 0).

Note: self.curr.bit_count() is a method that returns the number of set bits (ones) in self.curr, which is used to determine if the current bit configuration corresponds to the desired combinationLength.

Intuitively, we are navigating a binary tree, where each node represents a combination. Moving to the next combination is done by traversing the tree in a specific order, in this case, lexicographical order, which correlates to the decreasing numerical value of the binary representation of self.curr.

Learn more about Backtracking patterns.

Solution Approach

The CombinationIterator harnesses the efficiency of bit manipulation and bitwise operations to deal with combination generation. Here's an in-depth look into how the solution's implementation handles the process:

1. Data Structure: The class does not use an actual data structure like an array or list to store combinations. Instead, it relies on an integer (self.curr) to represent the combinations as sets of bits, with each bit corresponding to whether a character in the string is included (1) or not (2). The string characters are reversed in (self.cs) to facilitate the formation of combinations in lexicographical order when bits are right shifted during combination extraction.

2. Initialization:

   • self.curr is initialized to (1 << len(characters)) - 1, which sets all bits to 1, to the length of characters. This represents the full set of characters.
   • self.size holds the required combination length.
   • self.cs holds the reversed string of characters for easier iteration from least significant bit to most significant bit in the context of the combination string.

3. The next() Function:

   • The while loop decrements self.curr until it finds a number whose binary representation contains exactly self.size number of 1 bits.
   • The for loop iterates through the bits of self.curr. Whenever it encounters a set bit, it appends the corresponding character from self.cs to the ans list.
   • Finally, the current combination (ans) is updated by reversing and joining to form the string before self.curr is decremented to move to the next combination.

4. The hasNext() Function:

   • Similar to next(), hasNext() uses a while loop to decrement self.curr until the number of bits set matches self.size.
   • Then, it simply checks whether self.curr is non-negative, meaning there is at least one more combination remaining.

5. Algorithm: The algorithm efficiently generates the next combination on-the-fly without maintaining a collection of all combinations which could be computationally expensive. It uses a binary counter decreasing from the largest possible value and ensures only those binary numbers with the correct number of 1 bits are accepted as valid combinations.

6. Performance: This implementation exhibits both time and space efficiency. It does not compute all combinations upfront, thereby saving on space. The time complexity of the next() method is O(n), where n is the length of the characters string, and is based on the number of bits that need to be checked. The hasNext() method runs in the same complexity due to the similar while loop.

By utilizing bit manipulation and bitwise operations, the solution effectively iterates over the combinations in lexicographical order without pre-computing an extensive list of all possible combinations.

Example Walkthrough

Let's use a small example to illustrate the solution approach. Suppose we have the following input:

characters = "abc", and combinationLength = 2.

The CombinationIterator will produce combinations of the 2 letters from the 3 letter input string in lexicographical order.

Initialization:

• self.curr is initialized to (1 << 3) - 1, which is 7 in binary (111). This represents the full set of characters 'abc'.
• self.size is set to 2 because we want combinations of two characters.
• self.cs stores the reversed string "cba" so we start extracting combinations from right to left in terms of significance.

The next() Function:

• We call next(), the algorithm starts by checking if self.curr, which is 111 in binary or 7 in decimal, has exactly 2 bits set. In this case, it does, so it proceeds to create the next combination.
• The combination "ab" (110 in binary) is derived by including the first and second characters of the reversed string, which are "b" and "c" in the original string, thus forming the combination "bc".
• self.curr is decremented to 110 in binary or 6 in decimal after forming the first combination.

The hasNext() Function:

• After calling next() once, we can call hasNext() to see if there's another combination. Since self.curr is 6 (110 in binary) and can be decremented further to find another combination with two 1 bits, hasNext() will return true.

Subsequent Calls to next() and hasNext():

• We call next() again, and this time, self.curr = 6 (110). It has exactly 2 bits set, so it's a valid combination. The next combination, "ac" (101 in binary), is formed by including the first and third characters of the reversed string ("c" and "a" in the original string), resulting in "ca".
• self.curr is then decremented to 101 in binary or 5 in decimal.
• If we were to call hasNext() now, it would return true again since self.curr can still be decremented to the next valid combination.
• The next call to next() would decrement self.curr to 011 (3 in decimal), corresponding to the combination "bc" in the original string, which translates to "cb" in the reversed string.
• Once self.curr reaches a value that doesn't have 2 bits set after decrementing (in this case, when it's less than 3), calling next() would not yield any more combinations, and hasNext() would return false.

This process will iterate through all the combinations ("bc", "ca", "ab") without pre-calculated list and in a memory-efficient way, only computing the next combination when next() is called.

Solution Implementation

Time and Space Complexity

The given code defines a class CombinationIterator which is responsible for iterating over all combinations of a specific length from the given string. Here, we analyze the time complexity and space complexity for methods next and hasNext.

next method:

• Time Complexity: The worst-case time complexity of the next method can be considered as O(N) where N is the length of the characters string. This occurs because in the worst case, one might have to traverse all the bits to find the right combination that matches the required combinationLength. Besides, the construction of the combination itself takes O(N) in the worst case (where all characters are included in the combination).
• Space Complexity: The next method uses ans list to store the current combination, so in the worst case where the combination includes all characters, the space complexity will be O(N).

hasNext method:

• Time Complexity: The hasNext method internally performs the same process as the next method minus the part where the actual combination is constructed. Thus, it has a time complexity of O(N) for the same reasons as the next method since it potentially needs to check every bit of curr.
• Space Complexity: The hasNext method does not use any additional space that scales with the input size, hence the space complexity is O(1) as it updates the curr value only.

Overall:

The overall space complexity for the CombinationIterator class is O(N) due to the storage of the cs field which is the reversed characters string.

Note: bit_count() is a method that returns the number of set bits (1s) in the integer. As of Python 3.10, this is a built-in method, and its time complexity could be considered O(1) on average, though it could vary based on the implementation. If bit_count() is not available in an earlier Python version, an equivalent method to count bits would have a time complexity of O(log(curr)) due to counting the bits in a number; however, log(curr) is bounded by N, it will not be larger than N and therefore is negligible in comparison to the cost of traversing the string itself.

Learn more about how to find time and space complexity quickly using problem constraints.