Problem Description

You need to design a class called CombinationIterator that generates all combinations of a given length from a string of characters in lexicographical order.

The class should support three operations:

1. Constructor CombinationIterator(characters, combinationLength): Initialize the iterator with:

   • characters: a string of sorted, distinct lowercase English letters
   • combinationLength: the desired length of each combination

2. next(): Returns the next combination of the specified length in lexicographical order. Each call to this method should return a different combination until all combinations are exhausted.

3. hasNext(): Returns true if there are more combinations available, false otherwise.

For example, if you initialize the iterator with characters = "abc" and combinationLength = 2, the combinations would be "ab", "ac", "bc" in that order. Calling next() would return these combinations one by one, and hasNext() would return true until all combinations have been returned.

The solution uses a depth-first search (DFS) approach to generate all possible combinations during initialization. It explores two choices at each character position: either include the current character in the combination or skip it. When a combination reaches the desired length, it's added to the list. The class maintains an index to track which combination to return next, making the next() and hasNext() operations efficient with O(1) time complexity.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem doesn't involve nodes and edges or graph traversal. We're generating combinations from a string of characters.

Need to solve for kth smallest/largest?

• No: We're not finding a specific kth element. Instead, we need to generate all combinations in lexicographical order.

Involves Linked Lists?

• No: The problem works with strings and combinations, not linked list data structures.

Does the problem have small constraints?

• Yes: The problem involves generating combinations, which inherently has factorial-like growth. The number of combinations C(n, k) can quickly become large, but typically these problems have reasonable constraints since we need to enumerate all possibilities.

Brute force / Backtracking?

• Yes: We need to explore all possible combinations systematically. This is a classic backtracking scenario where we:
  • Make a choice (include or exclude a character)
  • Explore further with that choice
  • Backtrack when we've either found a valid combination or exhausted possibilities
  • Try the alternative choice

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This is the perfect pattern for generating all combinations because we need to systematically explore all possible selections of characters while maintaining the lexicographical order (which is naturally preserved when we process characters in their given sorted order).

Intuition

When we need to generate all combinations of a specific length from a set of characters, we face a classic selection problem: for each character, we must decide whether to include it or not in our current combination.

Think of it like building a team of k players from n available players. For each player, you make a binary decision - select them or skip them. This naturally leads to a recursive approach where we explore both possibilities at each step.

The key insight is that since our input characters are already sorted and we process them in order, the combinations will automatically be generated in lexicographical order. For example, with "abc" and length 2, we'll naturally get "ab" before "ac" before "bc" because we consider characters from left to right.

The backtracking pattern emerges from this decision tree:

1. Start with an empty combination
2. For each character position, we have two branches:
   • Include the current character (add it to our temporary combination)
   • Exclude the current character (skip to the next one)
3. When our combination reaches the target length, we've found a valid result
4. When we've processed all characters or can't possibly reach the target length, we backtrack

By pre-generating all combinations during initialization, we trade space for time - the next() and hasNext() operations become simple array lookups with O(1) complexity. This is practical because the iterator pattern implies we'll likely access most or all combinations, making the upfront computation worthwhile.

The recursive DFS naturally handles the "backtracking" by using the call stack. When we append a character, explore that path, then pop it off, we're essentially undoing our choice and trying the alternative - this is the essence of backtracking.

Learn more about Backtracking patterns.

Solution Approach

The solution implements a backtracking algorithm to generate all combinations during initialization, then provides simple iteration through the pre-computed results.

Data Structure Design:

• self.cs: A list storing all valid combinations in lexicographical order
• self.idx: An index pointer tracking the current position for iteration
• t: A temporary list used during DFS to build combinations character by character

The Core DFS Algorithm:

The dfs(i) function explores all possible combinations starting from index i:

1. Base Case - Valid Combination Found:

   if len(t) == combinationLength:
       cs.append(''.join(t))
       return

   When our temporary list t reaches the target length, we've found a valid combination. We convert it to a string and store it.

2. Base Case - Out of Bounds:

   if i == n:
       return

   If we've processed all characters without reaching the target length, we backtrack.

3. Recursive Exploration:

   t.append(characters[i])  # Choose to include current character
   dfs(i + 1)               # Explore with this choice
   t.pop()                  # Backtrack - undo the choice
   dfs(i + 1)               # Explore without including current character

   This is the heart of backtracking:

   • First, we include characters[i] in our combination
   • Recursively explore all combinations that include this character
   • Remove the character (backtrack) to restore the state
   • Explore all combinations that exclude this character

Why This Generates Lexicographical Order:

Since the input characters are sorted and we process them left-to-right, always trying "include" before "exclude", we naturally generate combinations in lexicographical order. For "abc" with length 2:

• First we try including 'a': leads to "ab", "ac"
• Then we try excluding 'a' but including 'b': leads to "bc"

Iterator Implementation:

Once all combinations are pre-computed:

• next(): Simply returns self.cs[self.idx] and increments the index - O(1) operation
• hasNext(): Checks if self.idx < len(self.cs) - O(1) operation

Time and Space Complexity:

• Initialization: O(C(n, k)) time and space, where n is the length of characters and k is combinationLength
• next() and hasNext(): O(1) time
• Total space: O(C(n, k)) to store all combinations

Example Walkthrough

Let's walk through the solution with characters = "abc" and combinationLength = 2.

Initialization Phase - Building All Combinations:

We start DFS from index 0 with an empty temporary list t = []:

DFS(0): t = []
├─ Include 'a': t = ['a']
│  └─ DFS(1): t = ['a']
│     ├─ Include 'b': t = ['a','b']
│     │  └─ DFS(2): len(t) = 2 ✓ → Store "ab"
│     └─ Exclude 'b': t = ['a']
│        └─ DFS(2): t = ['a']
│           ├─ Include 'c': t = ['a','c']
│           │  └─ DFS(3): len(t) = 2 ✓ → Store "ac"
│           └─ Exclude 'c': t = ['a']
│              └─ DFS(3): i = 3 (out of bounds) → return
└─ Exclude 'a': t = []
   └─ DFS(1): t = []
      ├─ Include 'b': t = ['b']
      │  └─ DFS(2): t = ['b']
      │     ├─ Include 'c': t = ['b','c']
      │     │  └─ DFS(3): len(t) = 2 ✓ → Store "bc"
      │     └─ Exclude 'c': t = ['b']
      │        └─ DFS(3): i = 3 (out of bounds) → return
      └─ Exclude 'b': t = []
         └─ DFS(2): t = []
            └─ Include 'c': t = ['c']
               └─ DFS(3): len(t) = 1 < 2, i = 3 → return

After initialization:

• self.cs = ["ab", "ac", "bc"]
• self.idx = 0

Using the Iterator:

1. hasNext() → returns True (0 < 3)
2. next() → returns "ab", increments idx to 1
3. hasNext() → returns True (1 < 3)
4. next() → returns "ac", increments idx to 2
5. hasNext() → returns True (2 < 3)
6. next() → returns "bc", increments idx to 3
7. hasNext() → returns False (3 ≮ 3)

The backtracking ensures we explore all paths: when we add 'a' and find combinations "ab" and "ac", we backtrack (remove 'a') to explore paths without 'a', finding "bc". The lexicographical order is maintained because we always process characters left-to-right and try "include" before "exclude".

Solution Implementation

Time and Space Complexity

Time Complexity:

• Initialization (__init__): O(C(n, k) * k) where n is the length of the characters string and k is the combinationLength. The DFS explores all possible combinations which is C(n, k) = n!/(k!(n-k)!), and for each valid combination, we join k characters together which takes O(k) time.

• next() method: O(1) - Simply returns the pre-computed combination at the current index and increments the index.

• hasNext() method: O(1) - Just compares two integers.

Space Complexity:

• Overall: O(C(n, k) * k) where n is the length of characters and k is the combinationLength.
  • The cs list stores all C(n, k) combinations, with each combination being a string of length k.
  • The recursion stack depth is at most O(n) since we traverse through all character positions.
  • The temporary list t uses at most O(k) space.
  • Therefore, the dominant factor is the storage of all combinations: O(C(n, k) * k).

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

Common Pitfalls

1. Memory Overflow with Large Inputs

The most significant pitfall is pre-computing and storing all combinations during initialization. For large inputs, this can cause memory issues. For example, with 26 characters and combinationLength of 13, there are C(26,13) = 10,400,600 combinations to store.

Solution: Use lazy evaluation with bit manipulation or direct computation:

class CombinationIterator:
    def __init__(self, characters: str, combinationLength: int):
        self.characters = characters
        self.combinationLength = combinationLength
        self.current = [i for i in range(combinationLength)]
        self.finished = False
      
    def next(self) -> str:
        result = ''.join(self.characters[i] for i in self.current)
        # Generate next combination indices
        n = len(self.characters)
        for i in range(self.combinationLength - 1, -1, -1):
            if self.current[i] < n - self.combinationLength + i:
                self.current[i] += 1
                for j in range(i + 1, self.combinationLength):
                    self.current[j] = self.current[j-1] + 1
                break
        else:
            self.finished = True
        return result
  
    def hasNext(self) -> bool:
        return not self.finished

2. Inefficient String Concatenation in Recursive Calls

Creating strings with ''.join(t) during recursion for every valid combination can be inefficient, especially if done repeatedly.

Solution: Store indices instead of actual strings, then generate strings on-demand:

def generate_combinations(index: int) -> None:
    if len(current_combination) == combinationLength:
        # Store indices instead of the actual string
        all_combinations.append(current_combination[:])
        return
    # ... rest of the logic

def next(self) -> str:
    indices = self.combinations[self.current_index]
    result = ''.join(self.characters[i] for i in indices)
    self.current_index += 1
    return result

3. Stack Overflow with Deep Recursion

For very long character strings with small combination lengths, the recursion depth could potentially exceed Python's default recursion limit (typically 1000).

Solution: Use iterative approach with explicit stack:

def generate_combinations_iterative(characters, combinationLength):
    result = []
    stack = [(0, [])]
  
    while stack:
        index, current = stack.pop()
      
        if len(current) == combinationLength:
            result.append(''.join(current))
            continue
          
        if index == len(characters):
            continue
      
        # Process in reverse order to maintain lexicographical order
        stack.append((index + 1, current[:]))  # Skip current character
        stack.append((index + 1, current + [characters[index]]))  # Include current
  
    return result[::-1]  # Reverse to get correct order

4. Not Handling Edge Cases

Failing to handle edge cases like empty strings or combinationLength greater than the string length.

Solution: Add validation in the constructor:

def __init__(self, characters: str, combinationLength: int):
    if not characters or combinationLength <= 0:
        self.combinations = []
        self.current_index = 0
        return
  
    if combinationLength > len(characters):
        self.combinations = []
        self.current_index = 0
        return
  
    # ... rest of initialization

5. Incorrect Assumption About Input Order

The code assumes the input characters are already sorted. If they're not, the output won't be in lexicographical order.

Solution: Sort the characters explicitly:

def __init__(self, characters: str, combinationLength: int):
    # Ensure characters are sorted for lexicographical order
    characters = ''.join(sorted(characters))
    # ... rest of initialization