# Find Permutation

Posted: 18 Feb, 2021
Difficulty: Moderate

## PROBLEM STATEMENT

#### If no such array exists, then you should return an empty array.

##### For example :
``````For N = 3 one valid array is [3,1,2,1,3,2].
``````
##### Input Format :
``````The first line of input contains an integer ‘T’, denoting the number of test cases. The test cases follow.

The first and the only line of each test case contains a single integer ‘N’.
``````
##### Output Format :
``````The checker will print “Valid” if the returned permutation is valid and follows all the conditions, otherwise, it will print “Invalid”. If an empty array is returned, the checker will print -1.

Print the output of each test case in a new line.
``````
##### Note :
``````You do not need to print anything. It has already been taken care of. Just implement the given function.
``````
##### Constraints :
``````1<= T <= 5
1 <= N <= 8

where ’T’ is the number of test cases and ‘N’  is the given integer.

Time Limit: 1 sec
`````` Approach 1

The idea is to make an array with exactly two occurrences of each element from 1 to N. Then we will generate all possible permutations of this array and check if any permutation is valid or not.

The steps are as follows:

1. Let’s define a recursive function as generatePermutations(arr, answer, N,  start, end), where arr is the array containing the permutation, answer is array that will store the final answer, ‘N’ is the given integer, ‘start’ is the starting index, and ‘end’ is the ending index. We will use this function to generate all possible permutations of the subarray of arr starting at index start and ending at index end.
2. Base condition: if ‘start’ is equal to ‘end’, then check if there are exactly k elements between both the occurrences of number ‘k’ for all valid ‘k’. (1<=k<=N). If it follows all the conditions, then make answer equal to the current array. Else ignore this array.
3. Iterate from i = start to end :
1. Swap the elements at index start, and i.
2. Call the recursive function again with left index as start+1, and right index as end i.e. generatePermutations(arr,start+1,end).
3. Swap back the elements at index start and i.