What is the time complexity of Count the Number of Square-Free Subsets?

The optimal solution has a time complexity of O(n * 2^10).

What is the space complexity of Count the Number of Square-Free Subsets?

The space complexity is O(2^10).

What data structures are used to solve Count the Number of Square-Free Subsets?

This problem uses Array, Math, Dynamic Programming, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count the Number of Square-Free Subsets?

Insight 1: The numbers in the input array are between 1 and 30. Since the square factors to avoid are 4, 9, and 25, any number divisible by one of those can't be part of a square-free subset with itself. This allows us to pre-compute which numbers are unsuitable and ignore them. Insight 2: We can represent each number in the array as a bitmask, where each bit corresponds to a prime number less than 30 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29). A set is square-free if no prime appears more than once in its product. Bitwise OR operations allow us to combine the prime factors of the numbers and keep track of which primes are already present.

Which companies ask LeetCode 2572. Count the Number of Square-Free Subsets in interviews?

This problem is frequently asked at Media.net.

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Count the Number of Square-Free Subsets - LeetCode 2572 Solution

Count the Number of Square-Free Subsets - Complete Solution Guide

Count the Number of Square-Free Subsets is LeetCode problem 2572, 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 positive integer 0-indexed array nums . A subset of the array nums is square-free if the product of its elements is a square-free integer . A square-free integer is an integer that is divisible by no square number other than 1 . Return the number of square-free non-empty subsets of the array nums . Since the answer may be too large, return it modulo 10 9 + 7 . A non-empty subset of nums is an array that can be obtained by deleting some (possibly none but not all) elements from nu

Detailed Explanation

The problem asks us to find the number of non-empty subsets of a given array `nums` such that the product of the elements in each subset is a square-free integer. A square-free integer is one that is not divisible by any perfect square other than 1. The elements of the array are positive integers between 1 and 30. We need to return the count of these square-free subsets modulo 10^9 + 7. For example, if `nums = [3, 4, 4, 5]`, the square-free subsets are `[3]`, `[5]`, and `[3, 5]`, giving a total of 3. If `nums = [1]`, the square-free subset is `[1]`, giving a total of 1.

Solution Approach

The solution uses dynamic programming and bit manipulation to efficiently count the number of square-free subsets. First, we precompute a bitmask for each number from 2 to 30, representing the prime factors of the number. If a number is divisible by 4, 9, or 25, we mark it as unsuitable and skip it. The dynamic programming table `dp` has `2^10` entries, where each entry `dp[mask]` stores the number of square-free subsets with the combined prime factors represented by the bitmask `mask`. The DP table is populated iteratively, considering each number in the input array. The number '1' is treated specially by multiplying the final result by 2 raised to the power of the number of 1's in the input array.

Step-by-Step Algorithm

Step 1: Initialize constants like MOD (10^9 + 7) and an array of primes PRIMES less than or equal to 30.
Step 2: Create a mapping `num_to_mask` to store the bitmask representation of each number from 2 to 30. Numbers divisible by 4, 9, or 25 are marked as unsuitable (-1).
Step 3: Count the occurrences of each number in the input array `nums` and store them in `counts` array.
Step 4: Initialize a DP table `dp` of size `2^(number of primes)`. `dp[0]` is set to 1, representing the empty subset.
Step 5: Iterate through numbers from 2 to 30. For each number, if it's suitable and present in the `counts` array, iterate through the DP table in reverse order.
Step 6: If the bitmask of the current number doesn't conflict with the current DP mask, update the DP table by adding the product of `dp[mask]` and the count of the current number to `dp[new_mask]`, where `new_mask` is the bitwise OR of `mask` and the bitmask of the current number. Take modulo MOD.
Step 7: Calculate the sum of all entries in the DP table (excluding dp[0] to avoid an empty set only containing 1's).
Step 8: Calculate 2 raised to the power of the number of 1's in the `nums` array. This accounts for all possible combinations of the number 1.
Step 9: Multiply the sum calculated in Step 7 by the result from Step 8 and take modulo MOD.
Step 10: Subtract 1 from the result (because problem statement does not want empty subset) and take modulo MOD to get the final answer.

Key Insights

Insight 1: The numbers in the input array are between 1 and 30. Since the square factors to avoid are 4, 9, and 25, any number divisible by one of those can't be part of a square-free subset with itself. This allows us to pre-compute which numbers are unsuitable and ignore them.
Insight 2: We can represent each number in the array as a bitmask, where each bit corresponds to a prime number less than 30 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29). A set is square-free if no prime appears more than once in its product. Bitwise OR operations allow us to combine the prime factors of the numbers and keep track of which primes are already present.
Insight 3: Dynamic Programming can be used to count the number of square-free subsets. The DP state can be represented by a bitmask indicating which prime factors are already used in the current subset. The number 1 needs special handling as it can be combined freely without affecting the square-free property.

Complexity Analysis

Time Complexity: O(n * 2^10)

Space Complexity: O(2^10)

Topics

This problem involves: Array, Math, Dynamic Programming, Bit Manipulation, Bitmask.

Companies

Asked at: Media.net.

Count the Number of Square-Free Subsets | LeetCode 2572

Medium

Array Math Dynamic Programming Bit Manipulation Bitmask

Time: O(n * 2^10)Space: O(2^10)

Solution Code

Available in:

class Solution {
    public int squareFreeSubsets(int[] nums) {
        int MOD = (int)1e9 + 7;
        int[] PRIMES = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
        int NUM_PRIMES = PRIMES.length;

        int[] num_to_mask = new int[31];
        for (int i = 0; i < 31; i++) {
            num_to_mask[i] = -1;
        }
        for (int i = 2; i < 31; i++) {
            if (i % 4 == 0 || i % 9 == 0 || i % 25 == 0) {
                continue;
            }

            int mask = 0;
            for (int p_idx = 0; p_idx < NUM_PRIMES; p_idx++) {
                int p = PRIMES[p_idx];
                if (i % p == 0) {
                    mask |= (1 << p_idx);
                }
            }
            num_to_mask[i] = mask;
        }

        int[] counts = new int[31];
        for (int num : nums) {
            counts[num]++;
        }

        int[] dp = new int[1 << NUM_PRIMES];
        dp[0] = 1;

        for (int num = 2; num < 31; num++) {
            if (counts[num] == 0) {
                continue;
            }

            int mask_num = num_to_mask[num];
            if (mask_num == -1) {
                continue;
            }

            int count = counts[num];
            for (int mask = (1 << NUM_PRIMES) - 1; mask >= 0; mask--) {
                if ((mask & mask_num) == 0) {
                    int new_mask = mask | mask_num;
                    dp[new_mask] = (int)(((long)dp[new_mask] + (long)dp[mask] * count) % MOD);
                }
            }
        }

        long total_subsets_gt1 = 0;
        for (int val : dp) {
            total_subsets_gt1 = (total_subsets_gt1 + val) % MOD;
        }

        int count_one = counts[1];
        long pow2_one = 1;
        for (int i = 0; i < count_one; i++) {
            pow2_one = (pow2_one * 2) % MOD;
        }

        long total_subsets = (total_subsets_gt1 * pow2_one) % MOD;

        long ans = (total_subsets - 1 + MOD) % MOD;

        return (int)ans;
    }
}

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Problem Info

Description

Problem 2572: Count the Number of Square-Free Subsets - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Bit Manipulation Bitmask

Companies

Media.net

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2572

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

10 steps

1
Step 1: Initialize constants like MOD (10^9 + 7) and an array of primes PRIMES less than or equal to 30.
2
Step 2: Create a mapping `num_to_mask` to store the bitmask representation of each number from 2 to 30. Numbers divisible by 4, 9, or 25 are marked as unsuitable (-1).
3
Step 3: Count the occurrences of each number in the input array `nums` and store them in `counts` array.
4
Step 4: Initialize a DP table `dp` of size `2^(number of primes)`. `dp[0]` is set to 1, representing the empty subset.
5
Step 5: Iterate through numbers from 2 to 30. For each number, if it's suitable and present in the `counts` array, iterate through the DP table in reverse order.
6
Step 6: If the bitmask of the current number doesn't conflict with the current DP mask, update the DP table by adding the product of `dp[mask]` and the count of the current number to `dp[new_mask]`, where `new_mask` is the bitwise OR of `mask` and the bitmask of the current number. Take modulo MOD.
7
Step 7: Calculate the sum of all entries in the DP table (excluding dp[0] to avoid an empty set only containing 1's).
8
Step 8: Calculate 2 raised to the power of the number of 1's in the `nums` array. This accounts for all possible combinations of the number 1.
9
Step 9: Multiply the sum calculated in Step 7 by the result from Step 8 and take modulo MOD.
10
Step 10: Subtract 1 from the result (because problem statement does not want empty subset) and take modulo MOD to get the final answer.

💡

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 * 2^10)

💾

Space Complexity

O(2^10)

📝

Why?

The time complexity is dominated by the nested loops in the dynamic programming portion. The outer loop iterates through numbers from 2 to 30 and the inner loop iterates through all possible subsets represented by the bitmasks (2^10). The space complexity is primarily due to the DP table, which requires O(2^10) space to store the number of square-free subsets for each possible combination of prime factors.

📖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

Recursion with Memoization

A recursive approach can also be used with memoization to avoid redundant calculations. The function will take the index of the current element and the current mask as input. If the product becomes non-square-free, we stop exploring the particular subset.

Time: O(n * 2^10)Space: O(n * 2^10)

Trade-offs: Recursion can be more intuitive to understand, but it might lead to stack overflow errors for larger inputs. Memoization is essential to achieve the same time complexity as the iterative DP approach.

✨

Key Insights

1Insight 1: The numbers in the input array are between 1 and 30. Since the square factors to avoid are 4, 9, and 25, any number divisible by one of those can't be part of a square-free subset with itself. This allows us to pre-compute which numbers are unsuitable and ignore them.
2Insight 2: We can represent each number in the array as a bitmask, where each bit corresponds to a prime number less than 30 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29). A set is square-free if no prime appears more than once in its product. Bitwise OR operations allow us to combine the prime factors of the numbers and keep track of which primes are already present.
3Insight 3: Dynamic Programming can be used to count the number of square-free subsets. The DP state can be represented by a bitmask indicating which prime factors are already used in the current subset. The number 1 needs special handling as it can be combined freely without affecting the square-free property.

📚

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.

Edge Cases

Edge case 1: The input array contains only the number 1. The solution correctly handles this by counting the number of 1's and multiplying the total count by 2 raised to the power of the number of 1's and subtract 1 since empty set is not allowed.
Edge case 2: The input array contains numbers divisible by perfect squares (4, 9, 25). The solution filters out these numbers in the `num_to_mask` mapping.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the number 1 separately. Since 1 doesn't affect the square-free property, it can be included or excluded from any subset, leading to 2^count_of_ones possible combinations.

✗Common mistake 2: Not taking modulo at each step of the calculations. The product of numbers can quickly exceed the integer limits, leading to incorrect results. Modulo operation should be applied after each multiplication and addition.

✗Common mistake 3: Not iterating in reverse order in the DP step. Modifying dp[mask | mask_num] based on the current value of dp[mask] before dp[mask] has been fully processed will count combinations multiple times, resulting in 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: Combinatorial optimization problems where certain combinations are invalid or restricted based on divisibility rules. For instance, selecting a set of components with constraints on shared resources.

🏦Real-world scenario 2: Cryptography, specifically in the selection of prime numbers for key generation. This concept is used to avoid keys with common factors, ensuring security.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: July 23, 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

❓

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 the Number of Square-Free 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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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