Leetcode #2572: Count the Number of Square-Free Subsets
In this guide, we solve Leetcode #2572 Count the Number of Square-Free Subsets in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Array, Math, Dynamic Programming, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: nums = [3,4,4,5]
Output: 3
Explanation: There are 3 square-free subsets in this example:
- The subset consisting of the 0th element [3]. The product of its elements is 3, which is a square-free integer.
- The subset consisting of the 3rd element [5]. The product of its elements is 5, which is a square-free integer.
- The subset consisting of 0th and 3rd elements [3,5]. The product of its elements is 15, which is a square-free integer.
It can be proven that there are no more than 3 square-free subsets in the given array.
Python Solution
class Solution:
def squareFreeSubsets(self, nums: List[int]) -> int:
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
cnt = Counter(nums)
mod = 10**9 + 7
n = len(primes)
f = [0] * (1 << n)
f[0] = pow(2, cnt[1])
for x in range(2, 31):
if cnt[x] == 0 or x % 4 == 0 or x % 9 == 0 or x % 25 == 0:
continue
mask = 0
for i, p in enumerate(primes):
if x % p == 0:
mask |= 1 << i
for state in range((1 << n) - 1, 0, -1):
if state & mask == mask:
f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod
return sum(v for v in f) % mod - 1
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.