Number of Squareful Arrays

update Feb 19 2019, 0:41

LeetCode

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 i such 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

Basic Idea:

Solution 1: DFS + Pruning

我们可以生成所有可能的permutation，在过程中不断检查新确认的数字和其之前的数字是否满足要求。如果不考虑pruning，时间复杂度为 O(N!)，剪枝之后复杂度显著降低，比DP更快。

• Java Code:

    class Solution {
        public int numSquarefulPerms(int[] A) {
            return dfs(A, 0);
        }
  
        private int dfs(int[] A, int pos) {
            if (pos == A.length) {
                return 1;
            }
            int ret = 0;
            // 用于去重，不用重复数字替换pos
            Set<Integer> used = new HashSet<>();
            for (int i = pos; i < A.length; ++i) {
                if (!used.add(A[i]) && i > pos) continue;
                if (pos > 0 && !isValid(A[i], A[pos - 1])) continue; // pruning
                swap(A, pos, i);
                ret += dfs(A, pos + 1);
                swap(A, pos, i);
            }
            return ret;
        }
  
        private void swap(int[] A, int i, int j) {
            int t = A[i];
            A[i] = A[j];
            A[j] = t;
        }
  
        private boolean isValid(int a, int b) {
            int sum = a + b;
            return (int)Math.pow((int)Math.sqrt(sum), 2) == sum;
        }
    }

  Solution 2: DP, Hamitonian Path

  花花酱的视频讲解：https://www.youtube.com/watch?v=ISnktonx7rY .

这里的实现比较复杂，因为涉及到的去重的操作比较复杂。