Number of squareful arrays

Posted: 13 Mar, 2021
Difficulty: Hard

PROBLEM STATEMENT

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You are given an Array/List Arr of non-negative integers. Your task is to return the number of Squareful permutations of the array.

An array is called Squareful if the sum of every pair of adjacent elements is a perfect square.

Example

ARR[1,3,6] is a Squareful array as 1+3=4 i.e. 2^2 , 3+6=9 i.e. 3^2.

Two permutations ARR1 and ARR2, are different from each other if there exit an index i such that ARR1[i] != ARR2[i].

Example:

[1,6,3] and [6,1,3] are different permutations.
Input format
The first line contains an integer ‘N’, representing the array’s size.

The second contains 'N' space-separated integers, denoting the elements of the array.
Output format
For each test case, return an integer denoting the number of Squareful Permutation of array ARR.
Note:
You don’t need to print anything or take input; it already has been taken care of. Just implement the function.
Constraints:-
1 <= N <= 12
0 <= ARR[i] <= 10^4 , where 0 <= i < N

Time limit: 1 sec
Approach 1

In this approach, we will check all the permutations of the array. And for each permutation in array ‘ARR’, check if ARR[i]+ARR[i+1] is a perfect square. If it is a perfect square, then increase the answer by 1. Return answer.

Algorithm:-

  1. Function for checking if a specific permutation of ‘ARR’ is Squareful or not.
  • Make a function bool CHECK(vector<int> &ARR)
  • Iterate ‘i’ from 0 to length of ‘ARR’.
    • If floor(sqrt( ARR[i] + ARR[i+1] ))!=ceil(sqrt(ARR[i] + ARR[i+1])), return false.
  • Else return false.
  1. Function for counting all Squareful permutation
  • Initialize an int ‘COUNT’ = 0;
  • Sort the array ‘ARR’ in increasing order.
  • Run a do-while loop to check all Squareful permutations.
  • do -> if CHECK(ARR) , then increase count.
  • While keep checking for the next permutation of ‘ARR’.
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