<p>An array is <strong>squareful</strong> if the sum of every pair of adjacent elements is a <strong>perfect square</strong>.</p>

<p>Given an integer array nums, return <em>the number of permutations of </em><code>nums</code><em> that are <strong>squareful</strong></em>.</p>

<p>Two permutations <code>perm1</code> and <code>perm2</code> are different if there is some index <code>i</code> such that <code>perm1[i] != perm2[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,17,8]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [1,8,17] and [17,8,1] are the valid permutations.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,2]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 12</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

'''
Change into graph problem.
Then just need to count num Hamiltonian paths (startin from any node)
That can be done with backtracking.
'''

squares = [
    i*i for i in range(31623)
]

class Solution:
    def numSquarefulPerms(self, nums: List[int]) -> int:
        n=len(nums)

        freq=defaultdict(int)
        for x in nums:
            freq[x] += 1
        arr=list(freq.keys())
        # construct the adj list
        g = defaultdict(list)
        for x in arr:
            for y in arr:
                if x+y in squares:
                    g[x].append(y)
        
        print('freq', freq)
        print(g)

        # do backtracking
        # visit each node freq[x] times

        ans=0
        pl=0
        vis=defaultdict(int)
        def dfs(u):
            nonlocal ans, pl
            # update
            pl += 1
            vis[u] += 1
            # check for done
            if pl == len(nums):
                ans += 1

            for nbr in g[u]:
                if vis[nbr] < freq[nbr]: # check if we can go here again
                    # visit it
                    dfs(nbr)
            pl -= 1
            vis[u] -= 1
        for start in arr:
            dfs(start)
        return ans
17900