Leetcode #1286: Iterator for Combination
In this guide, we solve Leetcode #1286 Iterator for Combination in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design the CombinationIterator class: CombinationIterator(string characters, int combinationLength) Initializes the object with a string characters of sorted distinct lowercase English letters and a number combinationLength as arguments. next() Returns the next combination of length combinationLength in lexicographical order.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, String, Backtracking, Iterator
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
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
Python Solution
class CombinationIterator:
def __init__(self, characters: str, combinationLength: int):
def dfs(i):
if len(t) == combinationLength:
cs.append(''.join(t))
return
if i == n:
return
t.append(characters[i])
dfs(i + 1)
t.pop()
dfs(i + 1)
cs = []
n = len(characters)
t = []
dfs(0)
self.cs = cs
self.idx = 0
def next(self) -> str:
ans = self.cs[self.idx]
self.idx += 1
return ans
def hasNext(self) -> bool:
return self.idx < len(self.cs)
# Your CombinationIterator object will be instantiated and called as such:
# obj = CombinationIterator(characters, combinationLength)
# param_1 = obj.next()
# param_2 = obj.hasNext()
Complexity
The time complexity is Exponential (worst case). The space complexity is O(depth).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.