How do you solve Range Module on LeetCode?

Range Module (LeetCode #715) 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 list of intervals allows for efficient merging and querying.

What is the brute force solution for Range Module?

The brute force approach for Range Module is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves maintaining a list of all tracked numbers. Whenever we add or remove a range, we update this list directly. This is simple but inefficient for larger ranges. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Range Module?

Range Module uses the following concepts: Design, Segment Tree, Ordered Set. The recommended patterns to study are: Interval Merging, Segment Trees.

What are the interview tips for Range Module?

Always clarify the requirements and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Range Module?

Failing to merge intervals correctly when adding ranges. Not considering edge cases when removing ranges, such as fully overlapping intervals.

#715

Range Module

Hard

DesignSegment TreeOrdered SetInterval MergingSegment Trees

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sorted list of intervals to efficiently manage overlapping ranges. By merging intervals when adding or removing ranges, we can keep track of disjoint intervals, making queries faster.

⚙️

Algorithm

4 steps

1Step 1: Maintain a list of intervals sorted by the starting point.
2Step 2: For addRange, merge overlapping intervals and insert the new range.
3Step 3: For removeRange, split intervals that overlap with the range to be removed.
4Step 4: For queryRange, check if the range is fully covered by existing intervals.

solution.py38 lines

1# Full working Python code
2class RangeModule:
3    def __init__(self):
4        self.intervals = []
5
6    def addRange(self, left: int, right: int) -> None:
7        new_intervals = []
8        i = 0
9        while i < len(self.intervals) and self.intervals[i][1] < left:
10            new_intervals.append(self.intervals[i])
11            i += 1
12        while i < len(self.intervals) and self.intervals[i][0] < right:
13            left = min(left, self.intervals[i][0])
14            right = max(right, self.intervals[i][1])
15            i += 1
16        new_intervals.append((left, right))
17        while i < len(self.intervals):
18            new_intervals.append(self.intervals[i])
19            i += 1
20        self.intervals = new_intervals
21
22    def queryRange(self, left: int, right: int) -> bool:
23        for start, end in self.intervals:
24            if start < right and left < end:
25                return start <= left and right <= end
26        return False
27
28    def removeRange(self, left: int, right: int) -> None:
29        new_intervals = []
30        for start, end in self.intervals:
31            if end <= left or start >= right:
32                new_intervals.append((start, end))
33            else:
34                if start < left:
35                    new_intervals.append((start, left))
36                if end > right:
37                    new_intervals.append((right, end))
38        self.intervals = new_intervals

ℹ

Complexity note: The time complexity is O(n) for addRange and removeRange because we may need to merge or split intervals, which requires iterating through the list of intervals. Querying is O(log n) due to binary search-like behavior when checking intervals.

1Using a sorted list of intervals allows for efficient merging and querying.
2Understanding how to manage overlapping intervals is crucial for this problem.

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