LeetCode #2572 — MEDIUM

Count the Number of Square-Free Subsets

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 10⁹ + 7.

A non-empty subset of nums is an array that can be obtained by deleting some (possibly none but not all) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Example 1:

Input: nums = [3,4,4,5]
Output: 3
Explanation: There are 3 square-free subsets in this example:
- The subset consisting of the 0^th element [3]. The product of its elements is 3, which is a square-free integer.
- The subset consisting of the 3^rd element [5]. The product of its elements is 5, which is a square-free integer.
- The subset consisting of 0^th and 3^rd 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.

Example 2:

Input: nums = [1]
Output: 1
Explanation: There is 1 square-free subset in this example:
- The subset consisting of the 0^th element [1]. The product of its elements is 1, which is a square-free integer.
It can be proven that there is no more than 1 square-free subset in the given array.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 30

Step 01

Brute Force Baseline

Problem summary: 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 109 + 7. A non-empty subset of nums is an array that can be obtained by deleting some (possibly none but not all) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming · Bit Manipulation

Example 1

[3,4,4,5]

Example 2

[1]

Core Insight

What unlocks the optimal approach

There are 10 primes before number 30.
Label primes from {2, 3, … 29} with {0,1, … 9} and let DP(i, mask) denote the number of subsets before index: i with the subset of taken primes: mask.
If the mask and prime factorization of nums[i] have a common prime, then it is impossible to add to the current subset, otherwise, it is possible.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2572: Count the Number of Square-Free Subsets
class Solution {
    public int squareFreeSubsets(int[] nums) {
        int[] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
        int[] cnt = new int[31];
        for (int x : nums) {
            ++cnt[x];
        }
        final int mod = (int) 1e9 + 7;
        int n = primes.length;
        long[] f = new long[1 << n];
        f[0] = 1;
        for (int i = 0; i < cnt[1]; ++i) {
            f[0] = (f[0] * 2) % mod;
        }
        for (int x = 2; x < 31; ++x) {
            if (cnt[x] == 0 || x % 4 == 0 || x % 9 == 0 || x % 25 == 0) {
                continue;
            }
            int mask = 0;
            for (int i = 0; i < n; ++i) {
                if (x % primes[i] == 0) {
                    mask |= 1 << i;
                }
            }
            for (int state = (1 << n) - 1; state > 0; --state) {
                if ((state & mask) == mask) {
                    f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod;
                }
            }
        }
        long ans = 0;
        for (int i = 0; i < 1 << n; ++i) {
            ans = (ans + f[i]) % mod;
        }
        ans -= 1;
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2572: Count the Number of Square-Free Subsets
func squareFreeSubsets(nums []int) (ans int) {
	primes := []int{2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
	cnt := [31]int{}
	for _, x := range nums {
		cnt[x]++
	}
	const mod int = 1e9 + 7
	n := 10
	f := make([]int, 1<<n)
	f[0] = 1
	for i := 0; i < cnt[1]; i++ {
		f[0] = f[0] * 2 % mod
	}
	for x := 2; x < 31; x++ {
		if cnt[x] == 0 || x%4 == 0 || x%9 == 0 || x%25 == 0 {
			continue
		}
		mask := 0
		for i, p := range primes {
			if x%p == 0 {
				mask |= 1 << i
			}
		}
		for state := 1<<n - 1; state > 0; state-- {
			if state&mask == mask {
				f[state] = (f[state] + f[state^mask]*cnt[x]) % mod
			}
		}
	}
	ans = -1
	for _, v := range f {
		ans = (ans + v) % mod
	}
	return
}

# Accepted solution for LeetCode #2572: Count the Number of Square-Free Subsets
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

// Accepted solution for LeetCode #2572: Count the Number of Square-Free Subsets
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2572: Count the Number of Square-Free Subsets
// class Solution {
//     public int squareFreeSubsets(int[] nums) {
//         int[] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
//         int[] cnt = new int[31];
//         for (int x : nums) {
//             ++cnt[x];
//         }
//         final int mod = (int) 1e9 + 7;
//         int n = primes.length;
//         long[] f = new long[1 << n];
//         f[0] = 1;
//         for (int i = 0; i < cnt[1]; ++i) {
//             f[0] = (f[0] * 2) % mod;
//         }
//         for (int x = 2; x < 31; ++x) {
//             if (cnt[x] == 0 || x % 4 == 0 || x % 9 == 0 || x % 25 == 0) {
//                 continue;
//             }
//             int mask = 0;
//             for (int i = 0; i < n; ++i) {
//                 if (x % primes[i] == 0) {
//                     mask |= 1 << i;
//                 }
//             }
//             for (int state = (1 << n) - 1; state > 0; --state) {
//                 if ((state & mask) == mask) {
//                     f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod;
//                 }
//             }
//         }
//         long ans = 0;
//         for (int i = 0; i < 1 << n; ++i) {
//             ans = (ans + f[i]) % mod;
//         }
//         ans -= 1;
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2572: Count the Number of Square-Free Subsets
function squareFreeSubsets(nums: number[]): number {
    const primes: number[] = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29];
    const cnt: number[] = Array(31).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    const mod: number = Math.pow(10, 9) + 7;
    const n: number = primes.length;
    const f: number[] = Array(1 << n).fill(0);
    f[0] = 1;
    for (let i = 0; i < cnt[1]; ++i) {
        f[0] = (f[0] * 2) % mod;
    }
    for (let x = 2; x < 31; ++x) {
        if (cnt[x] === 0 || x % 4 === 0 || x % 9 === 0 || x % 25 === 0) {
            continue;
        }
        let mask: number = 0;
        for (let i = 0; i < n; ++i) {
            if (x % primes[i] === 0) {
                mask |= 1 << i;
            }
        }
        for (let state = (1 << n) - 1; state > 0; --state) {
            if ((state & mask) === mask) {
                f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod;
            }
        }
    }
    let ans: number = 0;
    for (let i = 0; i < 1 << n; ++i) {
        ans = (ans + f[i]) % mod;
    }
    ans -= 1;
    return ans >= 0 ? ans : ans + mod;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + C × M)

Space

O(M)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.