Leetcode #2526: Find Consecutive Integers from a Data Stream
In this guide, we solve Leetcode #2526 Find Consecutive Integers from a Data Stream 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
For a stream of integers, implement a data structure that checks if the last k integers parsed in the stream are equal to value. Implement the DataStream class: DataStream(int value, int k) Initializes the object with an empty integer stream and the two integers value and k.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Queue, Hash Table, Counting, Data Stream
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["DataStream", "consec", "consec", "consec", "consec"]
[[4, 3], [4], [4], [4], [3]]
Output
[null, false, false, true, false]
Explanation
DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
dataStream.consec(4); // Only 1 integer is parsed, so returns False.
dataStream.consec(4); // Only 2 integers are parsed.
// Since 2 is less than k, returns False.
dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
// Since 3 is not equal to value, it returns False.
Python Solution
class DataStream:
def __init__(self, value: int, k: int):
self.val, self.k = value, k
self.cnt = 0
def consec(self, num: int) -> bool:
self.cnt = 0 if num != self.val else self.cnt + 1
return self.cnt >= self.k
# Your DataStream object will be instantiated and called as such:
# obj = DataStream(value, k)
# param_1 = obj.consec(num)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.