#996

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Number of Squareful Arrays

Count the number of squareful arrays from the given list of integers by checking adjacent pairs' sums for perfect squares.

array hash table math dynamic programming backtracking

Problem Statement

Given an integer array nums, return the number of distinct permutations of the array such that every adjacent pair of elements sums to a perfect square.

Two permutations are considered distinct if there exists an index where the elements are different. An array is considered squareful if for every pair of adjacent elements, their sum is a perfect square.

Examples

Example 1

Input: nums = [1,17,8]

Output: 2

[1,8,17] and [17,8,1] are the valid permutations.

Example 2

Input: nums = [2,2,2]

Output: 1

Example details omitted.

Constraints

1 <= nums.length <= 12
0 <= nums[i] <= 109

Solution Approach

Backtracking with Hash Set Lookup

This approach utilizes backtracking to generate all permutations of the array, while pruning invalid paths by checking if the sum of adjacent elements is a perfect square. A hash set ensures that duplicate numbers don't lead to redundant calculations.

Perfect Square Sum Precomputation

Before performing backtracking, precompute which sums of pairs of numbers are perfect squares. This reduces the complexity of checking the sum condition during the backtracking steps, making the algorithm more efficient.

Efficient Permutation Generation with Sorting

Sort the array initially to facilitate skipping over duplicate permutations. This ensures that the backtracking approach only explores unique permutations and avoids unnecessary recalculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is heavily dependent on the backtracking steps and the number of permutations generated. In the worst case, this can approach factorial time, i.e., O(n!). Precomputing the perfect square sums reduces the overhead, but the final complexity still depends on the problem size. Space complexity depends on storing the permutations and checking valid sums, which can be O(n).

What Interviewers Usually Probe

Backtracking expertise
Proficiency with pruning techniques
Efficiency in handling permutations and checking perfect square sums

Common Pitfalls or Variants

Common pitfalls

Misunderstanding how to efficiently prune the search space
Failing to handle duplicate numbers correctly
Incorrectly checking perfect square conditions for sums of adjacent elements

Follow-up variants

Using dynamic programming to optimize the backtracking process
Reducing space complexity by optimizing hash lookup operations
Exploring alternative approaches with bit manipulation to track visited elements

FAQ

What is a squareful array?

A squareful array is one where the sum of every adjacent pair of elements is a perfect square.

What is the time complexity of solving this problem?

The time complexity is typically O(n!) in the worst case due to the backtracking approach and permutation generation.

How do we check if the sum of two numbers is a perfect square?

You can check if the sum is a perfect square by calculating the square root of the sum and verifying if it is an integer.

What is the role of hash sets in this solution?

Hash sets are used to avoid duplicate permutations and to efficiently look up previously computed values during the backtracking process.

Can we use dynamic programming to solve this problem?

Yes, dynamic programming can optimize the backtracking approach by storing results of subproblems, though it might not always be necessary.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def numSquarefulPerms(self, nums: List[int]) -> int:
        n = len(nums)
        f = [[0] * n for _ in range(1 << n)]
        for j in range(n):
            f[1 << j][j] = 1
        for i in range(1 << n):
            for j in range(n):
                if i >> j & 1:
                    for k in range(n):
                        if (i >> k & 1) and k != j:
                            s = nums[j] + nums[k]
                            t = int(sqrt(s))
                            if t * t == s:
                                f[i][j] += f[i ^ (1 << j)][k]

        ans = sum(f[(1 << n) - 1][j] for j in range(n))
        for v in Counter(nums).values():
            ans //= factorial(v)
        return ans

Continue Practicing

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