What is the time complexity of Non-overlapping Intervals?

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

What data structures are used to solve Non-overlapping Intervals?

This problem uses Array, Dynamic Programming, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Non-overlapping Intervals?

Insight 1: Sorting the intervals by their end times allows us to greedily select intervals that end earliest, maximizing the number of non-overlapping intervals we can keep. Insight 2: The greedy approach of selecting intervals based on their end times leads to an optimal solution because it leaves the most room for subsequent non-overlapping intervals.

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Non-overlapping Intervals - LeetCode 435 Solution

Q: Which companies ask LeetCode 435. Non-overlapping Intervals in interviews?

This problem is frequently asked at Capital One, Instacart, J.P. Morgan, Snowflake, Verkada and 2 more companies.

Non-overlapping Intervals - Complete Solution Guide

Non-overlapping Intervals is LeetCode problem 435, 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

Given an array of intervals intervals where intervals[i] = [start i , end i ] , return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping . Note that intervals which only touch at a point are non-overlapping . For example, [1, 2] and [2, 3] are non-overlapping. Example 1: Input: intervals = [[1,2],[2,3],[3,4],[1,3]] Output: 1 Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping. Example 2: Input: intervals = [[1,2],

Detailed Explanation

The problem asks us to find the minimum number of intervals that need to be removed from a given set of intervals so that the remaining intervals are non-overlapping. Two intervals are considered non-overlapping if they do not share any common time. Note that intervals which only touch at a point are considered non-overlapping (e.g., [1, 2] and [2, 3]). The input is a list of intervals where each interval is represented by a pair of numbers [start, end]. The output is an integer representing the minimum number of intervals to remove.

Solution Approach

The solution employs a greedy approach. First, the intervals are sorted based on their end times. Then, the algorithm iterates through the sorted intervals, keeping track of the end time of the last non-overlapping interval selected. For each interval, if its start time is greater than or equal to the end time of the last selected interval, it means the current interval is non-overlapping and can be kept. The number of intervals kept is counted. Finally, the minimum number of intervals to remove is calculated by subtracting the number of intervals kept from the total number of intervals.

Step-by-Step Algorithm

Step 1: Sort the input `intervals` array based on the end times of the intervals in ascending order.
Step 2: Initialize `kept` to 1, as the first interval is always kept.
Step 3: Initialize `last_end` to the end time of the first interval (intervals[0][1]).
Step 4: Iterate through the remaining intervals from the second interval to the end of the array.
Step 5: For each interval, compare its start time with `last_end`. If the start time is greater than or equal to `last_end`, it means the current interval is non-overlapping.
Step 6: If the interval is non-overlapping, increment `kept` by 1 and update `last_end` to the end time of the current interval.
Step 7: After iterating through all the intervals, the minimum number of intervals to remove is the total number of intervals minus `kept`.

Key Insights

Insight 1: Sorting the intervals by their end times allows us to greedily select intervals that end earliest, maximizing the number of non-overlapping intervals we can keep.
Insight 2: The greedy approach of selecting intervals based on their end times leads to an optimal solution because it leaves the most room for subsequent non-overlapping intervals.
Insight 3: The problem can be reframed as finding the maximum number of non-overlapping intervals, and then subtracting that number from the total number of intervals.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Sorting.

Companies

Asked at: Capital One, Instacart, J.P. Morgan, Snowflake, Verkada, Yandex, Zoho.

Non-overlapping Intervals | LeetCode 435

Medium

Array Dynamic Programming Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }

        Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
        
        int kept = 1;
        int last_end = intervals[0][1];
        
        for (int i = 1; i < intervals.length; i++) {
            int start = intervals[i][0];
            int end = intervals[i][1];
            if (start >= last_end) {
                kept += 1;
                last_end = end;
            }
        }
        
        return intervals.length - kept;
    }
}

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Problem Info

Description

Problem 435: Non-overlapping Intervals - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Sorting

Companies

Capital OneInstacartJ.P. MorganSnowflakeVerkadaYandexZoho

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:435

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: Sort the input `intervals` array based on the end times of the intervals in ascending order.
2
Step 2: Initialize `kept` to 1, as the first interval is always kept.
3
Step 3: Initialize `last_end` to the end time of the first interval (intervals[0][1]).
4
Step 4: Iterate through the remaining intervals from the second interval to the end of the array.
5
Step 5: For each interval, compare its start time with `last_end`. If the start time is greater than or equal to `last_end`, it means the current interval is non-overlapping.
6
Step 6: If the interval is non-overlapping, increment `kept` by 1 and update `last_end` to the end time of the current interval.
7
Step 7: After iterating through all the intervals, the minimum number of intervals to remove is the total number of intervals minus `kept`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the sorting algorithm used. Space complexity is O(1) because no additional data structures that scale with the input size are used, beyond a few constant-size variables.

📖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

Dynamic Programming

A dynamic programming approach could be used where dp[i] represents the maximum number of non-overlapping intervals considering the first i intervals. However, it would also require sorting and would generally be less efficient.

Time: O(n^2) or O(n log n) depending on how it's optimizedSpace: O(n)

Trade-offs: Less efficient than the greedy approach. Not recommended for this problem due to the simpler and more efficient greedy solution.

Sorting by Start Time

Sorting by start time and iterating is possible, but requires a more complex comparison logic within the loop to track the minimum end time among overlapping intervals to maximize the number of non-overlapping intervals kept. This makes the greedy approach less intuitive and potentially more error-prone.

Time: O(n log n)Space: O(1)

Trade-offs: Less intuitive than sorting by end time, making the solution more complex.

✨

Key Insights

1Insight 1: Sorting the intervals by their end times allows us to greedily select intervals that end earliest, maximizing the number of non-overlapping intervals we can keep.
2Insight 2: The greedy approach of selecting intervals based on their end times leads to an optimal solution because it leaves the most room for subsequent non-overlapping intervals.
3Insight 3: The problem can be reframed as finding the maximum number of non-overlapping intervals, and then subtracting that number from the total number of intervals.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The solution returns 0 correctly, as there are no intervals to remove.
Edge case 2: All intervals are overlapping. The solution correctly identifies that only one interval can be kept, and removes the rest (n-1).
Edge case 3: Intervals with the same start and end times need to be handled correctly. The solution correctly sorts by end time and avoids counting overlapping instances.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Sorting by start time instead of end time. This can lead to an incorrect result because it doesn't prioritize intervals that end earliest, preventing the selection of the maximum possible number of non-overlapping intervals.

✗Common mistake 2: Not handling the edge case of an empty input array, leading to potential errors.

✗Common mistake 3: Failing to account for intervals that 'touch' at a point as non-overlapping, causing unnecessary removals.

✅

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: Scheduling problems, such as scheduling meetings or tasks on a single resource. The goal is to maximize the number of tasks that can be completed without any overlap, which is analogous to maximizing the number of non-overlapping intervals.

🏦Real-world scenario 2: Resource allocation. Imagine allocating time slots on a machine. Each interval represents a job to be done, and the goal is to allocate the maximum number of jobs without time conflicts. This is a direct application of the non-overlapping interval concept.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 15, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Non-overlapping Intervals"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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