What is the time complexity of Data Stream as Disjoint Intervals?

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

What data structures are used to solve Data Stream as Disjoint Intervals?

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

What are the key insights for solving Data Stream as Disjoint Intervals?

Insight 1: Use a boolean array to efficiently track the presence of each number within the given range (0 to 10000). This allows for quick checks when constructing intervals. Insight 2: Implement a 'dirty' flag to avoid recomputing the intervals unnecessarily. Only recalculate when a new number has been added since the last call to `getIntervals()`.

Which companies ask LeetCode 352. Data Stream as Disjoint Intervals in interviews?

This problem is frequently asked at Databricks.

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Data Stream as Disjoint Intervals - LeetCode 352 Solution

Data Stream as Disjoint Intervals - Complete Solution Guide

Data Stream as Disjoint Intervals is LeetCode problem 352, 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 a data stream input of non-negative integers a 1 , a 2 , ..., a n , summarize the numbers seen so far as a list of disjoint intervals. Implement the SummaryRanges class: SummaryRanges() Initializes the object with an empty stream. void addNum(int value) Adds the integer value to the stream. int[][] getIntervals() Returns a summary of the integers in the stream currently as a list of disjoint intervals [start i , end i ] . The answer should be sorted by start i . Example 1: Input ["SummaryR

Detailed Explanation

The problem requires you to maintain a summary of a stream of non-negative integers as a list of disjoint intervals. The `SummaryRanges` class needs to implement three functionalities: `SummaryRanges()` to initialize an empty stream, `addNum(int value)` to add an integer to the stream, and `getIntervals()` to return a list of disjoint intervals representing the numbers seen so far, sorted by the start of each interval. The integers are constrained to be between 0 and 10000, and there are constraints on the number of calls to `addNum` and `getIntervals`.

Solution Approach

The provided solution uses a boolean array `seen` to keep track of whether a number has been added to the stream. When `addNum(value)` is called, the corresponding entry in `seen` is set to `true`. The `getIntervals()` method iterates through the `seen` array, building intervals by finding consecutive `true` values. The 'dirty' flag optimizes the process by ensuring that the interval list is only recalculated when a new number is added. This approach leverages the constrained input range to provide an efficient solution.

Step-by-Step Algorithm

Step 1: Initialize a boolean array `seen` of size 10001 with all values set to `false`. Also, initialize a boolean flag `dirty` to `true` and a list to store the disjoint intervals.
Step 2: `addNum(value)`: Check if `seen[value]` is `false`. If it is, set `seen[value]` to `true` and set `dirty` to `true`.
Step 3: `getIntervals()`: Check the `dirty` flag. If it's `false`, return the previously calculated list of intervals.
Step 4: If `dirty` is `true`, clear the existing list of intervals.
Step 5: Iterate through the `seen` array from index 0 to 10000.
Step 6: If `seen[i]` is `true`, start a new interval with `start = i` and continue iterating to find the end of the interval.
Step 7: Set `j = i` and increment `j` as long as `j + 1 < 10001` and `seen[j + 1]` is `true`.
Step 8: The end of the interval is `j`. Add the interval `[start, j]` to the list of intervals.
Step 9: Set `i = j + 1` to skip over the just-processed interval.
Step 10: If `seen[i]` is `false`, simply increment `i` to check the next number.
Step 11: After iterating through the entire `seen` array, set `dirty` to `false` and return the list of intervals.

Key Insights

Insight 1: Use a boolean array to efficiently track the presence of each number within the given range (0 to 10000). This allows for quick checks when constructing intervals.
Insight 2: Implement a 'dirty' flag to avoid recomputing the intervals unnecessarily. Only recalculate when a new number has been added since the last call to `getIntervals()`.
Insight 3: Linear scan is sufficient since the constraints on value (0 to 10000) allow for a fixed-size array. More complex data structures like balanced trees could be used for a wider range, but are not necessary here.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Binary Search, Design, Ordered Set.

Companies

Asked at: Databricks.

Data Stream as Disjoint Intervals | LeetCode 352

Hard

Binary Search Design Ordered Set

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

Solution Code

Available in:

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

class SummaryRanges {

    private boolean[] seen;
    private boolean dirty;
    private List<int[]> intervals;

    public SummaryRanges() {
        seen = new boolean[10001];
        dirty = true;
        intervals = new ArrayList<>();
    }

    public void addNum(int value) {
        if (!seen[value]) {
            seen[value] = true;
            dirty = true;
        }
    }

    public int[][] getIntervals() {
        if (!dirty) {
            return intervals.toArray(new int[intervals.size()][]);
        }

        intervals = new ArrayList<>();
        int i = 0;
        while (i < 10001) {
            if (seen[i]) {
                int start = i;
                int j = i;
                while (j + 1 < 10001 && seen[j + 1]) {
                    j += 1;
                }
                intervals.add(new int[]{start, j});
                i = j + 1;
            } else {
                i += 1;
            }
        }

        dirty = false;
        return intervals.toArray(new int[intervals.size()][]);
    }
}

/**
 * Your SummaryRanges object will be instantiated and called as such:
 * SummaryRanges obj = new SummaryRanges();
 * obj.addNum(value);
 * int[][] param_2 = obj.getIntervals();
 */

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Problem Info

Description

Problem 352: Data Stream as Disjoint Intervals - Detailed explanation and solution approaches.

Topics

Binary Search Design Ordered Set

Companies

Databricks

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:352

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

11 steps

1
Step 1: Initialize a boolean array `seen` of size 10001 with all values set to `false`. Also, initialize a boolean flag `dirty` to `true` and a list to store the disjoint intervals.
2
Step 2: `addNum(value)`: Check if `seen[value]` is `false`. If it is, set `seen[value]` to `true` and set `dirty` to `true`.
3
Step 3: `getIntervals()`: Check the `dirty` flag. If it's `false`, return the previously calculated list of intervals.
4
Step 4: If `dirty` is `true`, clear the existing list of intervals.
5
Step 5: Iterate through the `seen` array from index 0 to 10000.
6
Step 6: If `seen[i]` is `true`, start a new interval with `start = i` and continue iterating to find the end of the interval.
7
Step 7: Set `j = i` and increment `j` as long as `j + 1 < 10001` and `seen[j + 1]` is `true`.
8
Step 8: The end of the interval is `j`. Add the interval `[start, j]` to the list of intervals.
9
Step 9: Set `i = j + 1` to skip over the just-processed interval.
10
Step 10: If `seen[i]` is `false`, simply increment `i` to check the next number.
11
Step 11: After iterating through the entire `seen` array, set `dirty` to `false` and return the list of intervals.

💡

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 time complexity is dominated by the linear scan of the `seen` array in the `getIntervals` method. The space complexity is due to the fixed-size `seen` array and the potential size of the `intervals` list.

📖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

Binary Search and Sorted List/Set

Instead of a boolean array, maintain a sorted list or set of the numbers seen so far. When `getIntervals` is called, iterate through the sorted list to construct intervals. Binary search can be used to efficiently insert elements into the sorted list.

Time: O(N log N) for addNum and O(N) for getIntervals, where N is the number of distinct numbers added.Space: O(N), where N is the number of distinct numbers added.

Trade-offs: This approach is more efficient if the range of numbers is very large but the number of distinct numbers added is relatively small. It avoids the fixed-size array limitation of the primary solution. If the range of possible integers is significantly larger than the number of calls, this approach is superior. It also works if there are no constraints to the integer range.

Interval Tree

Use an interval tree to store the intervals. When a new number is added, search the tree to see if it overlaps with any existing intervals. If it does, merge the intervals. Otherwise, insert a new interval.

Time: O(log N) for addNum and getIntervals, where N is the number of intervals.Space: O(N), where N is the number of intervals.

Trade-offs: This approach is generally more complex to implement, but it can be more efficient if there are many merges and the number of disjoint intervals is small compared to the size of the data stream. It's well-suited for scenarios with frequent merges and a relatively small number of intervals.

✨

Key Insights

1Insight 1: Use a boolean array to efficiently track the presence of each number within the given range (0 to 10000). This allows for quick checks when constructing intervals.
2Insight 2: Implement a 'dirty' flag to avoid recomputing the intervals unnecessarily. Only recalculate when a new number has been added since the last call to `getIntervals()`.
3Insight 3: Linear scan is sufficient since the constraints on value (0 to 10000) allow for a fixed-size array. More complex data structures like balanced trees could be used for a wider range, but are not necessary here.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Adding a number that extends an existing interval. The algorithm correctly merges the intervals by finding the consecutive `true` values.
Edge case 2: Adding a number that falls between two existing intervals. The algorithm handles this correctly by creating a new interval or merging existing ones when getIntervals is called.
Edge case 3: Adding duplicate numbers. The `addNum` function first checks if the number has already been seen, so duplicate numbers are ignored which is the intended behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to set the `dirty` flag to `true` when adding a new number, which can lead to `getIntervals` returning outdated results.

✗Common mistake 2: Not handling the case where a new number can merge two existing intervals into one.

✗Common mistake 3: Not correctly initializing the `seen` array or the `intervals` list in the constructor.

✗Common mistake 4: Failing to clear the existing `intervals` when recalculating in `getIntervals`.

✅

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: Network address allocation. When assigning IP address ranges to different entities, this algorithm can be used to track the allocated ranges and ensure that they are disjoint.

🏦Real-world scenario 2: Resource scheduling. This algorithm can be used to represent free time slots in a scheduling system. Adding a meeting would be analogous to adding a number to the stream. The free time slots are the intervals.

💡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

Binary search isn't just for sorted arrays. It can be applied to any monotonic function. Look for the "search space" in the problem.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Integer overflow in (left + right) / 2. Use left + (right - left) / 2 instead. Also watch for infinite loops.

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: November 20, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Data Stream as Disjoint Intervals"?

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

🤖 Need help with this problem?