Count Integers in Intervals

Design and implement a data structure to efficiently add intervals and count the total number of integers covered by them.

design segment tree ordered set

Problem Statement

You are tasked with implementing a data structure that manages intervals of integers. The data structure must support two operations: adding a new interval and counting how many integers are covered by the added intervals. A single interval [left, right] represents all integers x such that left <= x <= right.

The class you need to implement is CountIntervals. The add(left, right) function adds the interval [left, right], and the count() function returns the total number of distinct integers covered by all intervals added so far. The challenge is to implement this efficiently with the constraint that at most 105 operations will be performed on the data structure.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["CountIntervals", "add", "add", "count", "add", "count"] [[], [2, 3], [7, 10], [], [5, 8], []] Output [null, null, null, 6, null, 8]

Explanation CountIntervals countIntervals = new CountIntervals(); // initialize the object with an empty set of intervals. countIntervals.add(2, 3); // add [2, 3] to the set of intervals. countIntervals.add(7, 10); // add [7, 10] to the set of intervals. countIntervals.count(); // return 6 // the integers 2 and 3 are present in the interval [2, 3]. // the integers 7, 8, 9, and 10 are present in the interval [7, 10]. countIntervals.add(5, 8); // add [5, 8] to the set of intervals. countIntervals.count(); // return 8 // the integers 2 and 3 are present in the interval [2, 3]. // the integers 5 and 6 are present in the interval [5, 8]. // the integers 7 and 8 are present in the intervals [5, 8] and [7, 10]. // the integers 9 and 10 are present in the interval [7, 10].

Constraints

1 <= left <= right <= 109
At most 105 calls in total will be made to add and count.
At least one call will be made to count.

Solution Approach

Efficient Interval Addition

To efficiently add intervals, consider using a segment tree or an ordered set that can handle overlapping intervals. Each time a new interval is added, the structure must merge it with any overlapping or adjacent intervals, ensuring no redundant integers are counted. This operation should be optimized to handle large inputs.

Counting Distinct Integers

For counting distinct integers covered by the intervals, utilize the structure's ability to store ranges in a way that allows for fast querying. By maintaining intervals in a non-overlapping and sorted manner, the total number of covered integers can be calculated in an optimized way without explicitly storing all the integers.

Space and Time Complexity Considerations

The space and time complexity of this problem heavily depend on the data structure used. A segment tree can offer logarithmic time complexity for interval additions and queries, but it comes with increased space complexity. An ordered set approach may be simpler in terms of implementation but still requires careful management of overlapping intervals to ensure efficiency.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for both adding intervals and counting the number of integers depends on the chosen data structure. A segment tree allows for efficient merging of intervals in O(log n) time for each operation, while an ordered set might offer simpler implementation but still requires merging and querying operations that are logarithmic or linear in time complexity. Space complexity depends on the number of intervals and the data structure used, but it is generally O(n).

What Interviewers Usually Probe

Look for familiarity with segment trees and ordered sets as applicable data structures for range queries and dynamic intervals.
Assess the candidate's ability to optimize the data structure operations, especially handling overlapping intervals.
Evaluate how well the candidate manages the trade-offs between time complexity and space efficiency in the proposed solution.

Common Pitfalls or Variants

Common pitfalls

Failing to handle overlapping intervals correctly, leading to incorrect count results.
Inefficient data structure choices, resulting in excessive space or time complexity for the given constraints.
Not optimizing for the count operation, which can result in unnecessary recomputations or slow queries.

Follow-up variants

Implementing a dynamic range query system where intervals can be modified after addition.
Using a balanced binary search tree (BST) or other tree structures to handle dynamic interval additions and queries.
Expanding the problem to support removal of intervals, adding a new layer of complexity to the solution.

FAQ

What is the main challenge in the 'Count Integers in Intervals' problem?

The main challenge is efficiently managing overlapping intervals and counting distinct integers, which requires a data structure that can quickly add intervals and perform range queries.

How can segment trees be used in this problem?

Segment trees can be used to store intervals in a way that allows efficient merging of overlapping intervals and quick querying of the total number of distinct integers covered by all intervals.

What are some potential pitfalls in this problem?

Common pitfalls include incorrectly merging overlapping intervals, selecting inefficient data structures, and failing to optimize the count operation for large inputs.

What data structures are useful for solving the 'Count Integers in Intervals' problem?

Segment trees and ordered sets are the most useful data structures for efficiently handling interval addition and range queries in this problem.

How does GhostInterview help with interval problems like this?

GhostInterview provides structured guidance on selecting the right data structures, optimizing interval operations, and managing complexity, helping you avoid common mistakes and improve solution efficiency.

terminal

Solution

Solution 1: Segment Tree (Dynamic Opening)

According to the problem description, we need to maintain a set of intervals that supports adding intervals and querying operations. For adding intervals, we can use a segment tree to maintain the interval set.

class Node:
    __slots__ = ("left", "right", "l", "r", "mid", "v", "add")

    def __init__(self, l, r):
        self.left = None
        self.right = None
        self.l = l
        self.r = r
        self.mid = (l + r) // 2
        self.v = 0
        self.add = 0


class SegmentTree:
    def __init__(self):
        self.root = Node(1, int(1e9) + 1)

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

    def query(self, l, r, node=None):
        if node is None:
            node = self.root
        if l > r:
            return 0
        if node.l >= l and node.r <= r:
            return node.v
        self.pushdown(node)
        v = 0
        if l <= node.mid:
            v += self.query(l, r, node.left)
        if r > node.mid:
            v += self.query(l, r, node.right)
        return v

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

    def pushdown(self, node):
        if node.left is None:
            node.left = Node(node.l, node.mid)
        if node.right is None:
            node.right = Node(node.mid + 1, node.r)
        if node.add != 0:
            left, right = node.left, node.right
            left.add = node.add
            right.add = node.add
            left.v = left.r - left.l + 1
            right.v = right.r - right.l + 1
            node.add = 0


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

    def add(self, left, right):
        self.tree.modify(left, right, 1)

    def count(self):
        return self.tree.query(1, int(1e9))


# Your CountIntervals object will be instantiated and called as such:
# obj = CountIntervals()
# obj.add(left, right)
# param_2 = obj.count()

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