What is the time complexity of Divide Intervals Into Minimum Number of Groups?

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

What data structures are used to solve Divide Intervals Into Minimum Number of Groups?

This problem uses Array, Two Pointers, Greedy, Sorting, Heap (Priority Queue), Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Divide Intervals Into Minimum Number of Groups?

Insight 1: Sorting intervals by their start times allows us to process them in order, making it easier to determine if a new interval can be added to an existing group or if a new group needs to be created. Insight 2: A min-heap (priority queue) is crucial for efficiently tracking the end times of the intervals currently in use by each group. This allows us to quickly determine the earliest available end time among all existing groups.

Which companies ask LeetCode 2406. Divide Intervals Into Minimum Number of Groups in interviews?

This problem is frequently asked at Walmart Labs.

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Divide Intervals Into Minimum Number of Groups - LeetCode 2406 Solution

Q: How to solve Divide Intervals Into Minimum Number of Groups?

The solution uses a greedy approach combined with a min-heap. First, the intervals are sorted in ascending order based on their start times. Then, we iterate through the sorted intervals. For each interval, we check if it can be placed in an existing group (i.e., if its start time is greater than the earliest ending time among all groups which is stored in the heap). If it can, we remove the group with the earliest ending time from the heap and add the current interval to that group by pushing current interval's ending time. Otherwise, we need to create a new group by pushing current interval's ending time to the heap. The size of the heap at any point represents the number of groups currently in use. The final size of the heap is the minimum number of groups needed.

Divide Intervals Into Minimum Number of Groups - Complete Solution Guide

Divide Intervals Into Minimum Number of Groups is LeetCode problem 2406, 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 a 2D integer array intervals where intervals[i] = [left i , right i ] represents the inclusive interval [left i , right i ] . You have to divide the intervals into one or more groups such that each interval is in exactly one group, and no two intervals that are in the same group intersect each other. Return the minimum number of groups you need to make . Two intervals intersect if there is at least one common number between them. For example, the intervals [1, 5] and [5, 8] interse

Detailed Explanation

The problem asks us to divide a given set of intervals into the minimum number of groups. Each interval must belong to exactly one group, and no two intervals within the same group can overlap. Two intervals [a, b] and [c, d] overlap if they share at least one common number, i.e., if there exists a number x such that a <= x <= b and c <= x <= d. The goal is to determine the smallest number of such groups required to accommodate all intervals.

Solution Approach

The solution uses a greedy approach combined with a min-heap. First, the intervals are sorted in ascending order based on their start times. Then, we iterate through the sorted intervals. For each interval, we check if it can be placed in an existing group (i.e., if its start time is greater than the earliest ending time among all groups which is stored in the heap). If it can, we remove the group with the earliest ending time from the heap and add the current interval to that group by pushing current interval's ending time. Otherwise, we need to create a new group by pushing current interval's ending time to the heap. The size of the heap at any point represents the number of groups currently in use. The final size of the heap is the minimum number of groups needed.

Step-by-Step Algorithm

Step 1: Sort the intervals by their start times in ascending order.
Step 2: Initialize a min-heap to store the end times of intervals in each group.
Step 3: Iterate through the sorted intervals.
Step 4: For each interval, check if the min-heap is empty or if the current interval's start time is greater than the smallest end time in the heap (heap's root).
Step 5: If the condition in Step 4 is true, it means the current interval can be added to an existing group that ends earliest. Pop the earliest end time from the heap.
Step 6: Push the current interval's end time into the heap (either creating a new group or replacing an existing one).
Step 7: After processing all intervals, the number of groups is equal to the number of elements in the heap (heap size).
Step 8: Return the heap size.

Key Insights

Insight 1: Sorting intervals by their start times allows us to process them in order, making it easier to determine if a new interval can be added to an existing group or if a new group needs to be created.
Insight 2: A min-heap (priority queue) is crucial for efficiently tracking the end times of the intervals currently in use by each group. This allows us to quickly determine the earliest available end time among all existing groups.
Insight 3: The number of groups needed corresponds to the maximum depth (number of concurrent intervals) at any given point when scanning from left to right (based on sorted intervals).

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Greedy, Sorting, Heap (Priority Queue), Prefix Sum.

Companies

Asked at: Walmart Labs.

Divide Intervals Into Minimum Number of Groups | LeetCode 2406

Medium

Array Two Pointers Greedy Sorting Heap (Priority Queue)Prefix Sum

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

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

        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));

        PriorityQueue<Integer> minHeap = new PriorityQueue<>();

        for (int[] interval : intervals) {
            int start = interval[0];
            int end = interval[1];

            if (!minHeap.isEmpty() && start > minHeap.peek()) {
                minHeap.poll();
            }

            minHeap.offer(end);
        }

        return minHeap.size();
    }
}

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Problem Info

Description

Problem 2406: Divide Intervals Into Minimum Number of Groups - Detailed explanation and solution approaches.

Topics

Array Two Pointers Greedy Sorting Heap (Priority Queue)Prefix Sum

Companies

Walmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2406

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: Sort the intervals by their start times in ascending order.
2
Step 2: Initialize a min-heap to store the end times of intervals in each group.
3
Step 3: Iterate through the sorted intervals.
4
Step 4: For each interval, check if the min-heap is empty or if the current interval's start time is greater than the smallest end time in the heap (heap's root).
5
Step 5: If the condition in Step 4 is true, it means the current interval can be added to an existing group that ends earliest. Pop the earliest end time from the heap.
6
Step 6: Push the current interval's end time into the heap (either creating a new group or replacing an existing one).
7
Step 7: After processing all intervals, the number of groups is equal to the number of elements in the heap (heap size).
8
Step 8: Return the heap size.

💡

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(nlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the sorting step and the heap operations. The space complexity comes from the heap, which in the worst case can store all end times of the intervals.

📖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

Sweep Line Algorithm

Use a sweep line algorithm to track the start and end times of intervals. Increment a counter at each start time and decrement at each end time. The maximum value of the counter represents the minimum number of groups.

Time: O(n log n), due to sorting the start and end timesSpace: O(n), to store the start and end times

Trade-offs: The sweep line approach might be slightly more intuitive for some, but the heap-based approach is generally easier to implement correctly. The complexities are asymptotically the same.

✨

Key Insights

1Insight 1: Sorting intervals by their start times allows us to process them in order, making it easier to determine if a new interval can be added to an existing group or if a new group needs to be created.
2Insight 2: A min-heap (priority queue) is crucial for efficiently tracking the end times of the intervals currently in use by each group. This allows us to quickly determine the earliest available end time among all existing groups.
3Insight 3: The number of groups needed corresponds to the maximum depth (number of concurrent intervals) at any given point when scanning from left to right (based on sorted intervals).

📚

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.

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 (intervals array is empty). The solution correctly returns 0 as no groups are needed.
Edge case 2: No overlapping intervals. All intervals can be placed in a single group. The solution will result in a heap of size 1.
Edge case 3: Intervals with the same start time. The sorting ensures the processing order, and the heap handles overlaps appropriately.
Edge case 4: Intervals that are adjacent (e.g., [1,5], [5,8]). These are considered intersecting. The solution should correctly identify that these require different groups.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the intervals by start time. This leads to incorrect grouping decisions.

✗Common mistake 2: Not using a min-heap to track the earliest ending times. Using a regular list would result in O(n) lookups for the earliest ending time, increasing the overall time complexity.

✗Common mistake 3: Incorrectly handling the case where intervals are adjacent, mistakenly considering them non-overlapping.

✗Common mistake 4: Memory leaks. Forgetting to free memory allocated by malloc (especially in C).

✅

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 meetings in a conference room. Each interval represents a meeting's start and end time. The minimum number of groups is the minimum number of conference rooms needed to accommodate all meetings without conflicts.

🏦Real-world scenario 2: Resource allocation. Intervals represent time periods when a resource is required. The algorithm determines the minimum number of identical resources required to meet all demands without any overlaps. For example, server allocation for web requests.

💡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.

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 6, 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 "Divide Intervals Into Minimum Number of Groups"?

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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