How do you solve Iterator for Combination on LeetCode?

Iterator for Combination (LeetCode #1286) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding backtracking is crucial for generating combinations efficiently.

What is the brute force solution for Iterator for Combination?

The brute force approach for Iterator for Combination is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves generating all possible combinations of the given characters of the specified length and storing them. This allows us to iterate through the combinations in lexicographical order. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Iterator for Combination?

Iterator for Combination uses the following concepts: String, Backtracking, Design, Iterator. The recommended patterns to study are: Backtracking, Recursion.

What are the interview tips for Iterator for Combination?

Explain your thought process clearly while coding. Be prepared to discuss the trade-offs between time and space complexity. Practice generating combinations using backtracking before the interview.

What are common mistakes when solving Iterator for Combination?

Not handling the index correctly when generating combinations. Assuming all combinations need to be stored upfront.

#1286

Iterator for Combination

Medium

StringBacktrackingDesignIteratorBacktrackingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a backtracking approach to generate combinations on-the-fly instead of storing all combinations upfront. This saves space and allows us to generate combinations in lexicographical order directly.

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Algorithm

3 steps

1Step 1: Use a backtracking function to generate the next combination based on the current index.
2Step 2: Maintain a current combination string and an index to track the position in the characters string.
3Step 3: Implement next() and hasNext() to utilize the backtracking function.

solution.py28 lines

1class CombinationIterator:
2    def __init__(self, characters: str, combinationLength: int):
3        self.characters = characters
4        self.combinationLength = combinationLength
5        self.current = ''
6        self.index = 0
7        self.has_next = self._generate_next()
8
9    def _generate_next(self):
10        if self.index == len(self.characters):
11            return False
12        if len(self.current) == self.combinationLength:
13            return True
14        for i in range(self.index, len(self.characters)):
15            self.current += self.characters[i]
16            self.index = i + 1
17            if self._generate_next():
18                return True
19            self.current = self.current[:-1]
20        return False
21
22    def next(self) -> str:
23        result = self.current
24        self.has_next = self._generate_next()
25        return result
26
27    def hasNext(self) -> bool:
28        return self.has_next

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Complexity note: The time complexity is O(n) because we generate combinations on-the-fly without storing all of them. The space complexity is O(n) for the current combination string.

1Understanding backtracking is crucial for generating combinations efficiently.
2Recognizing the need for space optimization can lead to better solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.