Range Module

Design a RangeModule to track and query half-open intervals using segment trees or ordered sets.

Problem Statement

The task is to design a data structure that supports a RangeModule for tracking and querying ranges of numbers. A range is represented as a half-open interval [left, right), which includes all numbers x where left <= x < right. The module must support three operations: adding a range, removing a range, and querying whether a specific range is fully covered.

The RangeModule class should implement three methods: addRange(left, right), removeRange(left, right), and queryRange(left, right). These methods will allow for adding, removing, and querying ranges efficiently. Operations need to be optimized for performance, especially since the range boundaries can go up to 10^9, and there may be up to 10^4 calls.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["RangeModule", "addRange", "removeRange", "queryRange", "queryRange", "queryRange"] [[], [10, 20], [14, 16], [10, 14], [13, 15], [16, 17]] Output [null, null, null, true, false, true]

Explanation RangeModule rangeModule = new RangeModule(); rangeModule.addRange(10, 20); rangeModule.removeRange(14, 16); rangeModule.queryRange(10, 14); // return True,(Every number in [10, 14) is being tracked) rangeModule.queryRange(13, 15); // return False,(Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked) rangeModule.queryRange(16, 17); // return True, (The number 16 in [16, 17) is still being tracked, despite the remove operation)

Constraints

1 <= left < right <= 109
At most 104 calls will be made to addRange, queryRange, and removeRange.

Solution Approach

Use Sorted Set of Intervals

The main approach to solving this problem is to maintain a sorted set of disjoint intervals. This structure enables efficient insertion, removal, and range queries. By keeping intervals disjoint and sorted, the add, remove, and query operations can be performed efficiently.

Efficient Querying with Binary Search

To quickly determine if a range is fully covered, binary search can be used to find overlapping intervals in the sorted set. This ensures that querying a range’s coverage is fast, leveraging the sorted order to minimize unnecessary comparisons.

Time Complexity Optimization

By maintaining a set of disjoint intervals and using binary search, the operations addRange, removeRange, and queryRange can all be optimized. These operations take time proportional to the number of intervals in the set, leading to an efficient solution even with the large input constraints.

Complexity Analysis

Metric	Value
Time	Let
Space	O(A+R)

The time complexity for addRange and removeRange operations is O(A+R), where A is the number of intervals to add or remove, and R is the number of intervals in the current set. The queryRange operation has a time complexity of O(log N) due to the binary search in the sorted set, where N is the number of intervals.

What Interviewers Usually Probe

The candidate should demonstrate understanding of range operations and how segment trees or ordered sets can optimize range tracking.
Watch for solutions that optimize for time complexity, especially during range queries.
Look for insights into managing disjoint intervals and minimizing redundant work during add and remove operations.

Common Pitfalls or Variants

Common pitfalls

Mismanaging the disjoint intervals can lead to inefficient queries, making the solution suboptimal.
Not utilizing binary search or sorted data structures properly could cause unnecessary linear time operations for queryRange.
Adding or removing ranges without maintaining the disjoint property of intervals can result in incorrect coverage during queries.

Follow-up variants

Allow for dynamic resizing of the range boundaries.
Extend the problem to handle overlapping ranges more efficiently.
Modify the solution to handle interval modifications instead of simple add/remove operations.

FAQ

How does the Range Module work?

The Range Module works by maintaining a sorted set of disjoint intervals, allowing for efficient add, remove, and query operations on ranges.

What is the time complexity of the addRange operation?

The time complexity of the addRange operation is O(A+R), where A is the number of intervals being added and R is the number of current intervals.

What data structures are useful for solving the Range Module problem?

Segment trees and ordered sets are useful for efficiently tracking and querying ranges in the Range Module problem.

What is a half-open interval?

A half-open interval [left, right) includes all real numbers x where left <= x < right, meaning the left endpoint is included, but the right endpoint is not.

How does GhostInterview help with the Range Module problem?

GhostInterview helps by providing a structured approach to solving the problem, emphasizing efficient data structures and optimizing range operations.

terminal

Solution

Solution 1: Segment Tree

According to the problem description, we need to maintain a set of intervals, supporting operations of interval addition, deletion, and query. For the addition and deletion of intervals, we can use a segment tree to maintain the set of intervals.

class Node:
    __slots__ = ['left', 'right', 'add', 'v']

    def __init__(self):
        self.left = None
        self.right = None
        self.add = 0
        self.v = False


class SegmentTree:
    __slots__ = ['root']

    def __init__(self):
        self.root = Node()

    def modify(self, left, right, v, l=1, r=int(1e9), node=None):
        if node is None:
            node = self.root
        if l >= left and r <= right:
            if v == 1:
                node.add = 1
                node.v = True
            else:
                node.add = -1
                node.v = False
            return
        self.pushdown(node)
        mid = (l + r) >> 1
        if left <= mid:
            self.modify(left, right, v, l, mid, node.left)
        if right > mid:
            self.modify(left, right, v, mid + 1, r, node.right)
        self.pushup(node)

    def query(self, left, right, l=1, r=int(1e9), node=None):
        if node is None:
            node = self.root
        if l >= left and r <= right:
            return node.v
        self.pushdown(node)
        mid = (l + r) >> 1
        v = True
        if left <= mid:
            v = v and self.query(left, right, l, mid, node.left)
        if right > mid:
            v = v and self.query(left, right, mid + 1, r, node.right)
        return v

    def pushup(self, node):
        node.v = bool(node.left and node.left.v and node.right and node.right.v)

    def pushdown(self, node):
        if node.left is None:
            node.left = Node()
        if node.right is None:
            node.right = Node()
        if node.add:
            node.left.add = node.right.add = node.add
            node.left.v = node.add == 1
            node.right.v = node.add == 1
            node.add = 0


class RangeModule:
    def __init__(self):
        self.tree = SegmentTree()

    def addRange(self, left: int, right: int) -> None:
        self.tree.modify(left, right - 1, 1)

    def queryRange(self, left: int, right: int) -> bool:
        return self.tree.query(left, right - 1)

    def removeRange(self, left: int, right: int) -> None:
        self.tree.modify(left, right - 1, -1)


# Your RangeModule object will be instantiated and called as such:
# obj = RangeModule()
# obj.addRange(left,right)
# param_2 = obj.queryRange(left,right)
# obj.removeRange(left,right)

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