What is the time complexity of Interval List Intersections?

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

What data structures are used to solve Interval List Intersections?

This problem uses Array, Two Pointers, Line Sweep. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Interval List Intersections?

Insight 1: The lists are sorted, allowing us to use a two-pointer approach to efficiently iterate through them. Insight 2: The intersection of two intervals [a, b] and [c, d] is [max(a, c), min(b, d)] if max(a, c) <= min(b, d).

Which companies ask LeetCode 986. Interval List Intersections in interviews?

This problem is frequently asked at DoorDash, Mixpanel, Nuro, Verkada, Yandex.

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Interval List Intersections - LeetCode 986 Solution

Q: How to solve Interval List Intersections?

The solution uses a two-pointer approach. We initialize two pointers, `i` and `j`, to the beginning of `firstList` and `secondList`, respectively. We iterate through both lists, comparing the current intervals. For each pair of intervals, we calculate the intersection. If the intersection is not empty, we add it to the result. We then advance the pointer of the list whose interval ends earlier. This ensures that we don't miss any intersections and that we process the intervals in sorted order.

Q: What is the space complexity of Interval List Intersections?

The space complexity is O(min(m,n)).

Q: Which companies ask LeetCode 986. Interval List Intersections in interviews?

This problem is frequently asked at DoorDash, Mixpanel, Nuro, Verkada, Yandex.

Interval List Intersections - Complete Solution Guide

Interval List Intersections is LeetCode problem 986, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two lists of closed intervals, firstList and secondList , where firstList[i] = [start i , end i ] and secondList[j] = [start j , end j ] . Each list of intervals is pairwise disjoint and in sorted order . Return the intersection of these two interval lists . A closed interval [a, b] (with a <= b ) denotes the set of real numbers x with a <= x <= b . The intersection of two closed intervals is a set of real numbers that are either empty or represented as a closed interval. For examp

Detailed Explanation

The problem asks us to find the intersection of two lists of closed intervals. Both lists are already sorted and contain disjoint intervals. The intersection of two intervals is another interval representing the overlapping region between them. If there is no overlap, the intersection is empty. The input consists of two lists of intervals, `firstList` and `secondList`, where each interval is represented as a pair of integers [start, end]. The output should be a list of intervals representing the intersections of the intervals in the two input lists.

Solution Approach

The solution uses a two-pointer approach. We initialize two pointers, `i` and `j`, to the beginning of `firstList` and `secondList`, respectively. We iterate through both lists, comparing the current intervals. For each pair of intervals, we calculate the intersection. If the intersection is not empty, we add it to the result. We then advance the pointer of the list whose interval ends earlier. This ensures that we don't miss any intersections and that we process the intervals in sorted order.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `i` and `j`, to 0, representing the current index in `firstList` and `secondList`, respectively.
Step 2: Initialize an empty list `result` to store the intersection intervals.
Step 3: While `i` is less than the length of `firstList` and `j` is less than the length of `secondList`:
Step 4: Get the current intervals from `firstList` and `secondList`: `[start_i, end_i] = firstList[i]` and `[start_j, end_j] = secondList[j]`.
Step 5: Calculate the potential intersection interval: `low = max(start_i, start_j)` and `high = min(end_i, end_j)`.
Step 6: If `low <= high`, an intersection exists. Add the interval `[low, high]` to the `result` list.
Step 7: Compare `end_i` and `end_j`. If `end_i < end_j`, increment `i`. Otherwise, increment `j`.
Step 8: Return the `result` list.

Key Insights

Insight 1: The lists are sorted, allowing us to use a two-pointer approach to efficiently iterate through them.
Insight 2: The intersection of two intervals [a, b] and [c, d] is [max(a, c), min(b, d)] if max(a, c) <= min(b, d).
Insight 3: We need to advance the pointer of the list whose interval ends earlier. This is because the current interval in that list cannot intersect with any future intervals in the other list, as the intervals are disjoint and sorted.

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(min(m,n))

Topics

This problem involves: Array, Two Pointers, Line Sweep.

Companies

Asked at: DoorDash, Mixpanel, Nuro, Verkada, Yandex.

Interval List Intersections | LeetCode 986

Medium

Array Two Pointers Line Sweep

Time: O(m+n)Space: O(min(m,n))

Solution Code

Available in:

class Solution {
    public int[][] intervalIntersection(int[][] firstList, int[][] secondList) {
        List<int[]> result = new ArrayList<>();
        int i = 0, j = 0;

        while (i < firstList.length && j < secondList.length) {
            int start_i = firstList[i][0];
            int end_i = firstList[i][1];
            int start_j = secondList[j][0];
            int end_j = secondList[j][1];

            int low = Math.max(start_i, start_j);
            int high = Math.min(end_i, end_j);

            if (low <= high) {
                result.add(new int[]{low, high});
            }

            if (end_i < end_j) {
                i++;
            } else {
                j++;
            }
        }

        return result.toArray(new int[result.size()][]);
    }
}

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Problem Info

Description

Problem 986: Interval List Intersections - Detailed explanation and solution approaches.

Topics

Array Two Pointers Line Sweep

Companies

DoorDashMixpanelNuroVerkadaYandex

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:986

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

8 steps

1
Step 1: Initialize two pointers, `i` and `j`, to 0, representing the current index in `firstList` and `secondList`, respectively.
2
Step 2: Initialize an empty list `result` to store the intersection intervals.
3
Step 3: While `i` is less than the length of `firstList` and `j` is less than the length of `secondList`:
4
Step 4: Get the current intervals from `firstList` and `secondList`: `[start_i, end_i] = firstList[i]` and `[start_j, end_j] = secondList[j]`.
5
Step 5: Calculate the potential intersection interval: `low = max(start_i, start_j)` and `high = min(end_i, end_j)`.
6
Step 6: If `low <= high`, an intersection exists. Add the interval `[low, high]` to the `result` list.
7
Step 7: Compare `end_i` and `end_j`. If `end_i < end_j`, increment `i`. Otherwise, increment `j`.
8
Step 8: Return the `result` list.

💡

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(m+n)

💾

Space Complexity

O(min(m,n))

📝

Why?

The time complexity is O(m+n) because we increment either `i` or `j` in each iteration of the while loop, and the loop continues until one of the pointers reaches the end of its list. The space complexity is O(min(m,n)) because the maximum number of intervals in the result list cannot exceed the number of intervals in the shorter input 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

Brute Force

Iterate through all pairs of intervals (one from each list) and check for intersections.

Time: O(m*n), where m is the length of `firstList` and n is the length of `secondList`.Space: O(min(m,n)) in the worst case.

Trade-offs: This approach is less efficient than the two-pointer approach due to the nested loops. It's simpler to understand but not recommended for larger lists.

✨

Key Insights

1Insight 1: The lists are sorted, allowing us to use a two-pointer approach to efficiently iterate through them.
2Insight 2: The intersection of two intervals [a, b] and [c, d] is [max(a, c), min(b, d)] if max(a, c) <= min(b, d).
3Insight 3: We need to advance the pointer of the list whose interval ends earlier. This is because the current interval in that list cannot intersect with any future intervals in the other list, as the intervals are disjoint and sorted.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: One or both lists are empty. The two-pointer approach correctly handles this case because the while loop condition `i < len(firstList) and j < len(secondList)` will immediately terminate, returning an empty result.
Edge case 2: No intersections between any intervals in the two lists. The `if low <= high` condition ensures that only valid intersections are added to the result list. If no intersections exist, the result list remains empty and is correctly returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the condition `low <= high` correctly. This condition is crucial for ensuring that a valid intersection exists before adding it to the result. Failing to check this can lead to incorrect results.

✗Common mistake 2: Incrementing the wrong pointer. It is crucial to increment the pointer of the interval that ends earlier to ensure that the algorithm correctly finds all intersections. Incorrectly incrementing the pointers will skip over potential intersections.

✅

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 scheduling. Imagine two people are trying to schedule a meeting. Each person has a list of available time slots represented as intervals. Finding the intersection of these intervals can help determine the available times for the meeting.

🏦Real-world scenario 2: Resource allocation. Two systems might have different availability windows for a shared resource. Finding the intersection of these availability windows helps determine when the resource can be used by both systems.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: August 26, 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

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Interval List Intersections"?

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

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