What is the time complexity of Count Integers in Intervals?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Integers in Intervals?

This problem uses Design, Segment Tree, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Integers in Intervals?

Insight 1: The core challenge lies in efficiently merging overlapping intervals. A naive approach of iterating through all intervals for each `add` operation would be too slow. Insight 2: Maintaining a sorted list of disjoint intervals allows for binary search (or linear search if list is short) to quickly locate the insertion point and identify intervals that need merging.

Which companies ask LeetCode 2276. Count Integers in Intervals in interviews?

This problem is frequently asked at Databricks, LinkedIn.

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Count Integers in Intervals - LeetCode 2276 Solution

Count Integers in Intervals - Complete Solution Guide

Count Integers in Intervals is LeetCode problem 2276, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an empty set of intervals, implement a data structure that can: Add an interval to the set of intervals. Count the number of integers that are present in at least one interval. Implement the CountIntervals class: CountIntervals() Initializes the object with an empty set of intervals. void add(int left, int right) Adds the interval [left, right] to the set of intervals. int count() Returns the number of integers that are present in at least one interval. Note that an interval [left, right]

Detailed Explanation

The problem requires designing a data structure, `CountIntervals`, that efficiently manages a set of intervals and can count the total number of integers covered by these intervals. The data structure needs to support two main operations: 1. `add(left, right)`: Adds a new interval `[left, right]` to the existing set of intervals. If the new interval overlaps or is adjacent to existing intervals, they should be merged into a larger interval. 2. `count()`: Returns the total number of integers covered by all the intervals in the set, accounting for overlaps (i.e., each integer should be counted only once).

Solution Approach

The solution uses a sorted list (or vector) of disjoint intervals. When a new interval is added, the following steps are performed: 1. Find the correct position to insert the new interval using binary search (or linear search). 2. Check for overlapping/adjacent intervals to the left and right of the insertion point. 3. Merge overlapping/adjacent intervals into a single, larger interval. 4. Update the total count of covered integers by subtracting the lengths of the original intervals and adding the length of the merged interval. The `count()` operation simply returns the precomputed total count.

Step-by-Step Algorithm

Step 1: **Initialization**: Create an empty list/vector `intervals` to store disjoint intervals and an integer `total_count` to store the total number of integers covered (initialized to 0).
Step 2: **`add(left, right)`**: Find the correct insertion point `i` for the new interval `[left, right]` in the sorted `intervals` list.
Step 3: Check if the interval at `i-1` overlaps or is adjacent to `[left, right]`. If so, update `merged_left` and `merged_right` to encompass the overlapping intervals.
Step 4: Check if the intervals starting from index `i` overlap or are adjacent to the merged interval. If so, extend the `merged_right` and increment index `i`.
Step 5: Update the `total_count` by subtracting the length of the intervals that are going to be merged and adding the length of the new merged interval.
Step 6: Remove the merged intervals from the `intervals` list and insert the new merged interval at the correct position.
Step 7: **`count()`**: Return the `total_count`.

Key Insights

Insight 1: The core challenge lies in efficiently merging overlapping intervals. A naive approach of iterating through all intervals for each `add` operation would be too slow.
Insight 2: Maintaining a sorted list of disjoint intervals allows for binary search (or linear search if list is short) to quickly locate the insertion point and identify intervals that need merging.
Insight 3: Keeping track of the total count as intervals are added and merged avoids the need to recompute the count from scratch each time `count()` is called, improving performance.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Design, Segment Tree, Ordered Set.

Companies

Asked at: Databricks, LinkedIn.

Count Integers in Intervals | LeetCode 2276

Hard

Design Segment Tree Ordered Set

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class CountIntervals {

    private List<int[]> intervals;
    private int total_count;

    public CountIntervals() {
        intervals = new ArrayList<>();
        total_count = 0;
    }

    public void add(int left, int right) {
        int i = 0;
        while (i < intervals.size() && intervals.get(i)[0] < left) {
            i++;
        }

        int merged_left = left;
        int merged_right = right;
        int start_merge_idx = i;

        if (i > 0 && intervals.get(i - 1)[1] >= left - 1) {
            start_merge_idx = i - 1;
            int[] prev_interval = intervals.get(start_merge_idx);
            merged_left = prev_interval[0];
            merged_right = Math.max(right, prev_interval[1]);
        }

        int end_merge_idx = start_merge_idx;
        while (end_merge_idx < intervals.size() && intervals.get(end_merge_idx)[0] <= merged_right + 1) {
            merged_right = Math.max(merged_right, intervals.get(end_merge_idx)[1]);
            end_merge_idx++;
        }

        for (int k = start_merge_idx; k < end_merge_idx; k++) {
            total_count -= (intervals.get(k)[1] - intervals.get(k)[0] + 1);
        }

        total_count += (merged_right - merged_left + 1);

        List<int[]> newIntervals = new ArrayList<>();
        for (int k = 0; k < start_merge_idx; k++) {
            newIntervals.add(intervals.get(k));
        }
        newIntervals.add(new int[]{merged_left, merged_right});
        for (int k = end_merge_idx; k < intervals.size(); k++) {
            newIntervals.add(intervals.get(k));
        }
        intervals = newIntervals;
    }

    public int count() {
        return total_count;
    }
}

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Problem Info

Description

Problem 2276: Count Integers in Intervals - Detailed explanation and solution approaches.

Topics

Design Segment Tree Ordered Set

Companies

DatabricksLinkedIn

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2276

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: **Initialization**: Create an empty list/vector `intervals` to store disjoint intervals and an integer `total_count` to store the total number of integers covered (initialized to 0).
2
Step 2: **`add(left, right)`**: Find the correct insertion point `i` for the new interval `[left, right]` in the sorted `intervals` list.
3
Step 3: Check if the interval at `i-1` overlaps or is adjacent to `[left, right]`. If so, update `merged_left` and `merged_right` to encompass the overlapping intervals.
4
Step 4: Check if the intervals starting from index `i` overlap or are adjacent to the merged interval. If so, extend the `merged_right` and increment index `i`.
5
Step 5: Update the `total_count` by subtracting the length of the intervals that are going to be merged and adding the length of the new merged interval.
6
Step 6: Remove the merged intervals from the `intervals` list and insert the new merged interval at the correct position.
7
Step 7: **`count()`**: Return the `total_count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `add` operation's time complexity is dominated by the merging process, which can require iterating through a significant portion of the existing intervals in the worst case. The `count` operation is a simple lookup and takes constant time. The space complexity is linear because we store the intervals in a list/vector.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Segment Tree

A segment tree can be used to represent the intervals and efficiently track the covered integers. Each node in the segment tree represents a range of integers, and the tree can be updated to reflect the addition of new intervals. Querying the total count becomes a matter of traversing the tree.

Time: O(log N) for `add` and `count`, where N is the range of possible integer values (10^9 in this case).Space: O(N), where N is the range of possible integer values.

Trade-offs: Segment trees offer better theoretical time complexity for `add` compared to the sorted list approach in the worst-case scenario (O(log N) vs O(n)). However, the segment tree requires significantly more space because the size of the tree depends on the range of integers (10^9), whereas the sorted list approach only stores the actual intervals. The large range makes Segment Tree not very practical for this problem.

Ordered Set (Balanced Binary Search Tree)

Use a balanced binary search tree (e.g., a red-black tree) to store the intervals. The tree can be augmented to store the size of each subtree, allowing for efficient calculation of the total count. This approach avoids the need for explicit merging.

Time: O(log n) for `add` and O(log n) for `count`, where n is the number of intervals.Space: O(n), where n is the number of intervals.

Trade-offs: This approach can provide better performance than the sorted list approach in some cases, as both `add` and `count` operations have logarithmic time complexity. However, the implementation can be more complex.

✨

Key Insights

1Insight 1: The core challenge lies in efficiently merging overlapping intervals. A naive approach of iterating through all intervals for each `add` operation would be too slow.
2Insight 2: Maintaining a sorted list of disjoint intervals allows for binary search (or linear search if list is short) to quickly locate the insertion point and identify intervals that need merging.
3Insight 3: Keeping track of the total count as intervals are added and merged avoids the need to recompute the count from scratch each time `count()` is called, improving performance.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The new interval is completely disjoint from all existing intervals. In this case, the algorithm correctly inserts the new interval into the sorted list without merging and updates the total count.
Edge case 2: The new interval completely covers one or more existing intervals. The algorithm correctly merges all covered intervals into the new interval and updates the total count.
Edge case 3: Adding a very large interval that overlaps with almost all existing intervals. This is the worst-case scenario for time complexity, as the algorithm may need to iterate through and merge nearly all existing intervals.
Edge case 4: Adding adjacent intervals which will also be merged correctly

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the new interval overlaps with or is adjacent to existing intervals, leading to incorrect interval merging and count calculation.

✗Common mistake 2: Not updating the total count correctly after merging intervals, resulting in an inaccurate `count` result.

✗Common mistake 3: Failing to sort the intervals, which can lead to incorrect insertion and merging logic.

✗Common mistake 4: Incorrectly calculating the number of integers covered by an interval. Remember that the interval [left, right] covers `right - left + 1` integers.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: **Calendar Management**: This algorithm can be used to manage events in a calendar, where each interval represents the start and end time of an event. Adding a new event involves checking for overlaps with existing events and merging them if necessary. The `count` operation can be used to determine the total time occupied by events.

🏦Real-world scenario 2: **Resource Allocation**: The algorithm can be used to allocate resources, such as memory or disk space, to different tasks. Each interval represents a block of allocated resources. Adding a new task involves finding a suitable block of free resources and potentially merging it with existing free blocks. The `count` operation can be used to determine the total amount of allocated resources.

🎮Real-world scenario 3: **Data Compression**: Some data compression techniques involve identifying and merging overlapping or adjacent regions of data. This algorithm could be adapted for such scenarios.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Integers in Intervals"?

For this hard-level Design problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether segment tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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