How do you solve Number of Squareful Arrays on LeetCode?

Number of Squareful Arrays (LeetCode #996) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n!) time complexity and O(n) space complexity. Key insight: Understanding perfect squares is crucial for this problem.

What is the brute force solution for Number of Squareful Arrays?

The brute force approach for Number of Squareful Arrays is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach generates all possible permutations of the array and checks each one to see if it meets the squareful condition. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n!).

What are the interview tips for Number of Squareful Arrays?

Always clarify the constraints and edge cases before diving into the solution. Discuss your thought process and approach before writing code. Practice backtracking problems to get comfortable with recursion and state management.

What are common mistakes when solving Number of Squareful Arrays?

Not considering the frequency of numbers, leading to duplicate permutations. Failing to check the square condition during permutation construction.

#996

Number of Squareful Arrays

Hard

ArrayHash TableMathDynamic ProgrammingBacktrackingBit ManipulationBitmaskBacktrackingFrequency Map

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n!)
Space	O(n)	O(n)

💡

Intuition

Time O(n!)Space O(n)

The optimal solution uses backtracking with a frequency map to avoid generating duplicate permutations. It only checks valid adjacent pairs during the construction of the permutation, making it more efficient.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map of the numbers in the array.
2Step 2: Use backtracking to build permutations, ensuring each adjacent pair sums to a perfect square.
3Step 3: Count valid permutations as they are formed.

solution.py27 lines

1import math
2from collections import Counter
3
4
5def is_perfect_square(x):
6    return int(math.sqrt(x))**2 == x
7
8
9def backtrack(nums, path, count, freq):
10    if len(path) == len(nums):
11        count[0] += 1
12        return
13    for num in freq:
14        if freq[num] > 0:
15            if len(path) == 0 or is_perfect_square(path[-1] + num):
16                path.append(num)
17                freq[num] -= 1
18                backtrack(nums, path, count, freq)
19                path.pop()
20                freq[num] += 1
21
22
23def numSquarefulPerms(nums):
24    freq = Counter(nums)
25    count = [0]
26    backtrack(nums, [], count, freq)
27    return count[0]

ℹ

Complexity note: The time complexity remains O(n!) due to the nature of permutations, but we avoid generating duplicates, making it more efficient in practice. The space complexity is O(n) for the recursion stack and frequency map.

1Understanding perfect squares is crucial for this problem.
2Using a frequency map helps in managing duplicates efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.