996. Number of Squareful Arrays

Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square.

Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index isuch that A1[i] != A2[i].

 

Example 1:

Input: [1,17,8]
Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. 

Example 2:

Input: [2,2,2]
Output: 1

 

Note:

1. 1 <= A.length <= 12
2. 0 <= A[i] <= 1e9

 

Approach #1: Math. [Java]

class Solution {
    int ret = 0;
    Map<Integer, Integer> cntMap = new HashMap<>();
    Map<Integer, Set<Integer>> squareMap = new HashMap<>();
    
    public int numSquarefulPerms(int[] A) {
        for (int num : A) {
            if (!cntMap.containsKey(num)) {
                cntMap.put(num, 1);
                squareMap.put(num, new HashSet<Integer>());
            } else {
                cntMap.put(num, cntMap.get(num) + 1);
            }
        }
        
        for (int num1 : cntMap.keySet()) {
            for (int num2 : cntMap.keySet()) {
                double c = Math.sqrt(num1 + num2);
                if (c == Math.floor(c)) {
                    squareMap.get(num1).add(num2);
                    squareMap.get(num2).add(num1);
                }
            }
        }
        
        for (int key : cntMap.keySet()) {
            dfs(key, A.length - 1);
        }
        
        return ret;
    }
    
    public void dfs(int key, int left) {
        cntMap.put(key, cntMap.get(key)-1);
        if (left == 0) ret++;
        else {
            for (int next : squareMap.get(key)) {
                if (cntMap.get(next) != 0) {
                    dfs(next, left - 1);
                }
            }
        }
        cntMap.put(key, cntMap.get(key)+1);
    }
}

　　

Reference:

https://leetcode.com/problems/number-of-squareful-arrays/discuss/238562/C%2B%2BPython-Backtracking