What is the time complexity of The Number of Beautiful Subsets?

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

What data structures are used to solve The Number of Beautiful Subsets?

This problem uses Array, Hash Table, Math, Dynamic Programming, Backtracking, Sorting, Combinatorics. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving The Number of Beautiful Subsets?

Insight 1: Grouping by Remainder: Numbers with the same remainder when divided by `k` are independent. If two numbers have different remainders modulo k, their absolute difference can never equal k. This allows us to process each remainder group separately. Insight 2: Dynamic Programming within Remainder Groups: For each remainder group, we can further divide the numbers into 'chains' where each number in a chain is k apart from the next. We use a dynamic programming approach to count the beautiful subsets within each chain: considering either including or excluding a number from the chain

Which companies ask LeetCode 2597. The Number of Beautiful Subsets in interviews?

This problem is frequently asked at Infosys.

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The Number of Beautiful Subsets - LeetCode 2597 Solution

The Number of Beautiful Subsets - Complete Solution Guide

The Number of Beautiful Subsets is LeetCode problem 2597, 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 an array nums of positive integers and a positive integer k . A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k . Return the number of non-empty beautiful subsets of the array nums . A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums . Two subsets are different if and only if the chosen indices to delete are different. Example 1: Input: nums = [2,4,6], k = 2 Output: 4 Explanati

Detailed Explanation

The problem asks us to find the number of non-empty 'beautiful' subsets from a given array `nums`. A subset is considered 'beautiful' if no two numbers within the subset have an absolute difference equal to `k`. We need to count all such subsets, excluding the empty subset. For instance, if `nums` is [2, 4, 6] and `k` is 2, [2], [4], [6], and [2, 6] are beautiful subsets, resulting in a count of 4. The constraints are that the array size is up to 18, and number values and k are up to 1000. Subsets are unique based on the indices of the selected elements from the original array.

Solution Approach

The solution uses a grouping strategy based on remainders modulo `k` to reduce the problem's complexity. It groups numbers with the same remainder together because numbers with different remainders cannot violate the 'beautiful' subset condition. Within each remainder group, it identifies 'chains' of numbers where each number is k apart from the next (e.g., if k=2, a chain might be 2, 4, 6). For each chain, it applies a dynamic programming-like approach to calculate the number of beautiful subsets. It calculates the total number of beautiful subsets across all groups and then subtracts 1 to exclude the empty set.

Step-by-Step Algorithm

Step 1: Group Numbers by Remainder: Iterate through `nums` and group numbers based on their remainder when divided by `k`. Use a hash map (or dictionary) to store these groups, where the key is the remainder and the value is another hash map containing the number and its frequency.
Step 2: Iterate Through Remainder Groups: Loop through each remainder group.
Step 3: Sort Numbers in Remainder Group: Sort the numbers within the remainder group to easily identify chains.
Step 4: Identify and Process Chains: Iterate through the sorted numbers. If a number hasn't been visited, start a new chain. Extend the chain by adding numbers that are `k` greater than the current number and are present in the group. Mark numbers in the chain as visited.
Step 5: Calculate Beautiful Subsets for Each Chain: Use a DP approach within a chain. Maintain two variables: `prev_skip` (number of beautiful subsets without including the current number) and `prev_take` (number of beautiful subsets including the current number). For each number in the chain, calculate the new `skip_i` (previous skip + previous take) and `take_i` (number of ways to take current element * prev_skip). Update `prev_skip` and `prev_take`.
Step 6: Multiply Chain Results: Multiply the number of beautiful subsets for each chain within the remainder group to get `group_res`.
Step 7: Multiply Group Results: Multiply the number of beautiful subsets for each remainder group to get `total_count`.
Step 8: Subtract 1: Subtract 1 from `total_count` to exclude the empty subset.

Key Insights

Insight 1: Grouping by Remainder: Numbers with the same remainder when divided by `k` are independent. If two numbers have different remainders modulo k, their absolute difference can never equal k. This allows us to process each remainder group separately.
Insight 2: Dynamic Programming within Remainder Groups: For each remainder group, we can further divide the numbers into 'chains' where each number in a chain is k apart from the next. We use a dynamic programming approach to count the beautiful subsets within each chain: considering either including or excluding a number from the chain
Insight 3: Bit Manipulation for Counting Subsets: Using the expression (1 << count) -1 efficiently calculates the number of ways to take elements from a chain. This is because each element can either be included or excluded, giving 2^count possibilities. Subtracting 1 excludes the case where no elements are included (empty chain subset).

Complexity Analysis

Time Complexity: O(n*n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Math, Dynamic Programming, Backtracking, Sorting, Combinatorics.

Companies

Asked at: Infosys.

The Number of Beautiful Subsets | LeetCode 2597

Medium

Array Hash Table Math Dynamic Programming Backtracking Sorting Combinatorics

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int beautifulSubsets(int[] nums, int k) {
        Map<Integer, Map<Integer, Integer>> groups = new HashMap<>();
        for (int num : nums) {
            int rem = num % k;
            if (!groups.containsKey(rem)) {
                groups.put(rem, new HashMap<>());
            }
            Map<Integer, Integer> group = groups.get(rem);
            group.put(num, group.getOrDefault(num, 0) + 1);
        }

        int total_count = 1;

        for (int rem : groups.keySet()) {
            Map<Integer, Integer> group = groups.get(rem);
            List<Integer> sorted_nums = new ArrayList<>(group.keySet());
            Collections.sort(sorted_nums);

            int group_res = 1;
            Set<Integer> visited = new HashSet<>();

            for (int num : sorted_nums) {
                if (visited.contains(num)) {
                    continue;
                }

                List<Integer> chain = new ArrayList<>();
                int curr = num;
                while (group.containsKey(curr)) {
                    chain.add(curr);
                    visited.add(curr);
                    curr += k;
                }

                int prev_skip = 1;
                int prev_take = 0;

                for (int chain_num : chain) {
                    int count = group.get(chain_num);
                    int ways_to_take = (1 << count) - 1;

                    int skip_i = prev_skip + prev_take;
                    int take_i = ways_to_take * prev_skip;

                    prev_skip = skip_i;
                    prev_take = take_i;
                }

                int chain_res = prev_skip + prev_take;
                group_res *= chain_res;
            }

            total_count *= group_res;
        }

        return total_count - 1;
    }
}

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Problem Info

Description

Problem 2597: The Number of Beautiful Subsets - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Dynamic Programming Backtracking Sorting Combinatorics

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2597

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: Group Numbers by Remainder: Iterate through `nums` and group numbers based on their remainder when divided by `k`. Use a hash map (or dictionary) to store these groups, where the key is the remainder and the value is another hash map containing the number and its frequency.
2
Step 2: Iterate Through Remainder Groups: Loop through each remainder group.
3
Step 3: Sort Numbers in Remainder Group: Sort the numbers within the remainder group to easily identify chains.
4
Step 4: Identify and Process Chains: Iterate through the sorted numbers. If a number hasn't been visited, start a new chain. Extend the chain by adding numbers that are `k` greater than the current number and are present in the group. Mark numbers in the chain as visited.
5
Step 5: Calculate Beautiful Subsets for Each Chain: Use a DP approach within a chain. Maintain two variables: `prev_skip` (number of beautiful subsets without including the current number) and `prev_take` (number of beautiful subsets including the current number). For each number in the chain, calculate the new `skip_i` (previous skip + previous take) and `take_i` (number of ways to take current element * prev_skip). Update `prev_skip` and `prev_take`.
6
Step 6: Multiply Chain Results: Multiply the number of beautiful subsets for each chain within the remainder group to get `group_res`.
7
Step 7: Multiply Group Results: Multiply the number of beautiful subsets for each remainder group to get `total_count`.
8
Step 8: Subtract 1: Subtract 1 from `total_count` to exclude the empty subset.

💡

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*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Time Complexity: The algorithm iterates through each number in nums (O(n)) to group them. Within each group, it sorts the numbers O(n log n) in the worst case. Then for each sorted number it iterates (in worst case n elements and it checks for presence of each element in groups) so the time complexity of building the 'chain' is O(n). Finally the DP part for chain elements takes O(n) in worst case, so overall complexity is O(n + nlogn + n^2), which can be written as O(n^2) because n^2 dominates for larger n. The C implementation, using qsort, results in O(n log n) sorting complexity. The nested loops to build chains and calculate subsets contribute to an overall complexity of close to O(n*n) in the worst case when k is small or all elements have the same remainder. Space Complexity: The space complexity is dominated by the hash maps used to store the groups, which can store at most n key-value pairs, so overall space complexity is O(n).

📖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

Backtracking

A backtracking approach could explore all possible subsets and check if they are beautiful. This involves recursively including or excluding each number. If a subset is not beautiful, backtrack.

Time: O(2^n * n)Space: O(n)

Trade-offs: Backtracking has an exponential time complexity, making it suitable only for very small input sizes. It's generally less efficient than the provided grouping and DP-like approach. The provided solution is much more suitable due to the problem constraints.

✨

Key Insights

1Insight 1: Grouping by Remainder: Numbers with the same remainder when divided by `k` are independent. If two numbers have different remainders modulo k, their absolute difference can never equal k. This allows us to process each remainder group separately.
2Insight 2: Dynamic Programming within Remainder Groups: For each remainder group, we can further divide the numbers into 'chains' where each number in a chain is k apart from the next. We use a dynamic programming approach to count the beautiful subsets within each chain: considering either including or excluding a number from the chain
3Insight 3: Bit Manipulation for Counting Subsets: Using the expression (1 << count) -1 efficiently calculates the number of ways to take elements from a chain. This is because each element can either be included or excluded, giving 2^count possibilities. Subtracting 1 excludes the case where no elements are included (empty chain subset).

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

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

Edge Cases

Edge case 1: Empty Input Array: The code handles empty input array gracefully because the initial total_count is 1, and after the loops finish, it subtracts 1 returning 0 which is the correct answer.
Edge case 2: k = 1: When k is 1, no two numbers can exist together in a 'beautiful' subset. The grouping logic still works correctly, resulting in each number being in its own chain, and the calculation correctly accounts for single-element subsets only.
Edge case 3: Duplicate Numbers: The code correctly handles duplicate numbers using the frequency counts stored in the hash maps, calculating the number of ways to take combinations of these duplicate numbers into subsets.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly Calculating Subsets: Forgetting to subtract 1 to exclude the empty subset is a common error.

✗Common mistake 2: Ignoring Duplicate Numbers: Failing to account for the frequency of numbers when calculating subsets leads to an undercount.

✗Common mistake 3: Incorrect Grouping: Not understanding the key insight of grouping by remainder leads to trying all subset combinations, and exponential time complexity.

✅

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: Recommendation Systems: In recommendation systems, this approach can be adapted to avoid recommending items that are too similar based on a certain 'similarity' threshold (represented by 'k'). We want to find a diverse set of items for recommendation.

🏦Real-world scenario 2: Resource Allocation: Allocating resources such that no two allocated resources have a 'conflict' level greater than 'k'. For instance, assigning communication frequencies to different devices such that interference is minimized.

💡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 17, 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 13

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

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "The Number of Beautiful Subsets"?

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

🤖 Need help with this problem?