How do you solve Data Stream as Disjoint Intervals on LeetCode?

Data Stream as Disjoint Intervals (LeetCode #352) 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: Using a sorted data structure helps maintain order and allows for efficient merging of intervals.

What is the brute force solution for Data Stream as Disjoint Intervals?

The brute force approach for Data Stream as Disjoint Intervals is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves maintaining a list of all numbers seen so far and then checking for intervals by iterating through this list. This is straightforward but inefficient, especially as the number of elements increases. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Data Stream as Disjoint Intervals?

Always clarify the requirements, especially around edge cases. Think about the data structures that can help optimize your solution. Discuss your thought process and any trade-offs with the interviewer.

What are common mistakes when solving Data Stream as Disjoint Intervals?

Failing to handle duplicates properly. Not considering the merging of intervals when adding new numbers.

#352

Data Stream as Disjoint Intervals

Hard

Hash TableBinary SearchUnion-FindDesignData StreamOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a data structure that allows for efficient insertion and interval merging. By using a sorted set, we can quickly find where to insert new numbers and merge intervals as necessary.

⚙️

Algorithm

3 steps

1Step 1: Use a sorted data structure (like a TreeSet or SortedSet) to store the numbers.
2Step 2: When adding a number, check adjacent numbers to see if it merges with existing intervals.
3Step 3: Create intervals based on the merged ranges and return them.

solution.py37 lines

1class SummaryRanges:
2    def __init__(self):
3        self.intervals = []
4
5    def addNum(self, value: int) -> None:
6        if not self.intervals:
7            self.intervals.append([value, value])
8            return
9        for i in range(len(self.intervals)):
10            start, end = self.intervals[i]
11            if value < start:
12                self.intervals.insert(i, [value, value])
13                break
14            elif start <= value <= end:
15                return
16            elif value == end + 1:
17                self.intervals[i][1] = value
18                if i + 1 < len(self.intervals) and self.intervals[i + 1][0] == value + 1:
19                    self.intervals[i][1] = self.intervals[i + 1][1]
20                    self.intervals.pop(i + 1)
21                return
22            elif value == start - 1:
23                self.intervals[i][0] = value
24                return
25            elif value > end:
26                if i + 1 < len(self.intervals) and self.intervals[i + 1][0] == value + 1:
27                    self.intervals[i][1] = self.intervals[i + 1][1]
28                    self.intervals.pop(i + 1)
29                elif i + 1 == len(self.intervals):
30                    self.intervals.append([value, value])
31                else:
32                    continue
33        else:
34            self.intervals.append([value, value])
35
36    def getIntervals(self) -> List[List[int]]:
37        return self.intervals

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the set of numbers once to create intervals, and the space complexity is O(n) due to storing the intervals.

1Using a sorted data structure helps maintain order and allows for efficient merging of intervals.
2Handling edge cases like duplicates and adjacent numbers is crucial.

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