What is the time complexity of Count Ways to Group Overlapping Ranges?

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

What data structures are used to solve Count Ways to Group Overlapping Ranges?

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

What are the key insights for solving Count Ways to Group Overlapping Ranges?

Insight 1: Overlapping ranges effectively form connected components. All ranges within a connected component must belong to the same group. Insight 2: Sorting the ranges based on their start values simplifies the process of identifying overlapping ranges and merging them into connected components.

Which companies ask LeetCode 2580. Count Ways to Group Overlapping Ranges in interviews?

This problem is frequently asked at IBM.

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Count Ways to Group Overlapping Ranges - LeetCode 2580 Solution

Count Ways to Group Overlapping Ranges - Complete Solution Guide

Count Ways to Group Overlapping Ranges is LeetCode problem 2580, 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 ranges where ranges[i] = [start i , end i ] denotes that all integers between start i and end i (both inclusive ) are contained in the i th range. You are to split ranges into two (possibly empty) groups such that: Each range belongs to exactly one group. Any two overlapping ranges must belong to the same group. Two ranges are said to be overlapping if there exists at least one integer that is present in both ranges. For example, [1, 3] and [2, 5] are overlapping

Detailed Explanation

The problem asks us to find the number of ways to divide a given set of ranges into two groups, with the constraint that overlapping ranges must belong to the same group. The result needs to be calculated modulo 10^9 + 7. Each range in the input 'ranges' is defined by a start and end integer, [start_i, end_i], representing all integers within that interval. The output is the total number of ways to group the ranges satisfying the overlapping constraint.

Solution Approach

The solution approach involves first sorting the ranges by their start values. Then, we iterate through the sorted ranges, merging overlapping ranges into connected components. We keep track of the rightmost endpoint seen so far. If the current range's start is greater than the current maximum end, then it belongs to a new component. Finally, the number of ways to group the ranges is 2 raised to the power of the number of connected components, modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Sort the input array 'ranges' based on the start values of each range in ascending order.
Step 2: Initialize 'num_components' to 0 and 'current_max_end' to -1. 'num_components' will keep track of the number of connected components, and 'current_max_end' will store the largest endpoint encountered so far in the current component.
Step 3: Iterate through the sorted 'ranges'. For each range [start, end]:
Step 4: Check if 'start' is greater than 'current_max_end'. If it is, it means the current range does not overlap with the previous ranges, so increment 'num_components'.
Step 5: Update 'current_max_end' to be the maximum of its current value and 'end'. This extends the current component to include the new range if it overlaps or updates to the new end.
Step 6: After iterating through all ranges, calculate 2^'num_components' modulo (10^9 + 7). This is the total number of ways to group the ranges.
Step 7: Return the calculated result.

Key Insights

Insight 1: Overlapping ranges effectively form connected components. All ranges within a connected component must belong to the same group.
Insight 2: Sorting the ranges based on their start values simplifies the process of identifying overlapping ranges and merging them into connected components.
Insight 3: Each connected component can be assigned to either group 1 or group 2. Therefore, if there are 'n' connected components, there are 2^n possible groupings.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(1)

Topics

This problem involves: Array, Sorting.

Companies

Asked at: IBM.

Count Ways to Group Overlapping Ranges | LeetCode 2580

Medium

Array Sorting

Time: O(nlogn)Space: O(1)

Solution Code

Available in:

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

class Solution {
    public int countWays(int[][] ranges) {
        int MOD = (int)1e9 + 7;
        
        if (ranges == null || ranges.length == 0) {
            return 1;
        }
        
        Arrays.sort(ranges, Comparator.comparingInt(a -> a[0]));
        
        int num_components = 0;
        int current_max_end = -1;
        
        for (int[] range : ranges) {
            int start = range[0];
            int end = range[1];
            
            if (start > current_max_end) {
                num_components += 1;
            }
            
            current_max_end = Math.max(current_max_end, end);
        }
        
        long result = 1;
        for (int i = 0; i < num_components; i++) {
            result = (result * 2) % MOD;
        }
        
        return (int) result;
    }
}

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Problem Info

Description

Problem 2580: Count Ways to Group Overlapping Ranges - Detailed explanation and solution approaches.

Topics

Array Sorting

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2580

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 array 'ranges' based on the start values of each range in ascending order.
2
Step 2: Initialize 'num_components' to 0 and 'current_max_end' to -1. 'num_components' will keep track of the number of connected components, and 'current_max_end' will store the largest endpoint encountered so far in the current component.
3
Step 3: Iterate through the sorted 'ranges'. For each range [start, end]:
4
Step 4: Check if 'start' is greater than 'current_max_end'. If it is, it means the current range does not overlap with the previous ranges, so increment 'num_components'.
5
Step 5: Update 'current_max_end' to be the maximum of its current value and 'end'. This extends the current component to include the new range if it overlaps or updates to the new end.
6
Step 6: After iterating through all ranges, calculate 2^'num_components' modulo (10^9 + 7). This is the total number of ways to group the ranges.
7
Step 7: Return the calculated result.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dictated by the sorting algorithm applied on the 'ranges' array. Common efficient sorting algorithms, like merge sort or quicksort (used internally by `Arrays.sort` in Java and `std::sort` in C++), have a time complexity of O(n log n). The space complexity remains constant because the algorithm modifies the input array in place and uses a fixed number of 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

Disjoint Set Union (DSU)

Use Disjoint Set Union (also known as Union-Find) data structure to find connected components. Iterate through each pair of ranges and if they overlap, union them using DSU. The number of disjoint sets at the end gives the number of components.

Time: O(n^2 * α(n)) or close to O(n^2), where n is the number of ranges and α(n) is the Inverse Ackermann function, which grows extremely slowly and is practically constant. The n^2 factor comes from comparing each pair of ranges. Using path compression and union by rank optimizations, the DSU operations are nearly constant time.Space: O(n) to store the parent array for the DSU data structure.

Trade-offs: The DSU approach might be conceptually simpler to understand for some, but it has a higher time complexity (O(n^2)) compared to the sorting-based approach (O(n log n)). Therefore, the sorting-based approach is generally preferred for larger inputs.

✨

Key Insights

1Insight 1: Overlapping ranges effectively form connected components. All ranges within a connected component must belong to the same group.
2Insight 2: Sorting the ranges based on their start values simplifies the process of identifying overlapping ranges and merging them into connected components.
3Insight 3: Each connected component can be assigned to either group 1 or group 2. Therefore, if there are 'n' connected components, there are 2^n possible groupings.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: Empty input 'ranges'. The solution correctly handles this case by returning 1 (2^0 = 1), as there's only one way to group zero ranges: both groups are empty.
Edge case 2: Non-overlapping ranges. The algorithm correctly identifies each range as a separate component, resulting in 2^n groupings, where n is the number of ranges.
Edge case 3: All ranges overlap. The algorithm will identify a single component, resulting in 2^1 = 2 groupings (all ranges in either group 1 or group 2).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the ranges. Without sorting, it's difficult to efficiently identify overlapping ranges, leading to incorrect results.

✗Common mistake 2: Integer overflow when calculating 2^'num_components'. For larger values of 'num_components', the result can exceed the maximum value of an integer, so using modulo arithmetic is crucial.

✗Common mistake 3: Assuming ranges are always sorted or non-empty without explicit checks in the code.

✗Common mistake 4: Incorrectly determining overlapping condition. Ranges [a, b] and [c, d] overlap if max(a,c) <= min(b,d). Confusing this condition leads to wrong number of components and incorrect results.

✅

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: Task scheduling. Ranges can represent time intervals for different tasks. Grouping tasks such that overlapping tasks are scheduled on the same resource (e.g., a CPU core).

🏦Real-world scenario 2: Network routing. Ranges can represent network address blocks. Grouping address blocks for efficient routing configurations, where overlapping blocks must be routed through the same path.

💡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: July 31, 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 13

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

@AlgoMasterJul 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.

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 "Count Ways to Group Overlapping Ranges"?

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

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