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Positive AND Codehef October long challenge Solution |CodeChef October long challenge editorial

Positive AND codechef October long challenge Solution |Codechef October long challenge editorial -

This Problem is taken from Codechef  October long challenge and I have explained Positive and complete solution.

Problem Statement-

A permutation p1,p2...pN of {1,2,...,N} is beautiful if pi&pi+1 is greater than 0 for every 1i<N . You are given an integer N, and your task is toconstruct a beautiful permutation of length N or determine that it's impossible.

Note that a&b denotes the bitwise AND of a and b.

Input:

First-line will contain T, number of testcases. Then the testcases follow. Each testcase contains a single line of input, an integer N.

Output:

For each test case output 1 if there is no suitable permutation of length N, otherwise output N integers in a single line which form a beautiful permutation. If there are multiple answers output any of them.

Constraints

  • 1N105
  • The sum of N over all test cases does not exceed 106

Subtasks

  • 50 points : 1N,T9
  • 50 points : Original constraints

Sample Input:

3
4
3
5

Sample Output:

-1
1 3 2
2 3 1 5 4
CodeChef long challenge editorial





Solution-

Positive AND problem was little bit tricky problem.
mistake 1- Do not try to build the same output as given in custom input.
mistake 2- these can more then one ans of this problem so you can print any one of them.

lets jump to my approach
I was thinking while solving the problem what wrong with the given permutation why it is not valid solution.
so i got ans by the observation that when a 2^n(1,2,4,8,16....) number AND with 2^n-1(1,3,7,15) number it forms always zero because in 2^n number only MSB present(eg 2=10,4=100,8=1000) other bit are equals to zero.
So my assumed permutation can be valid answer if i swap 2^n number with there next elements. so that it can not form 0 while doing AND 
operation.
To make my permutation valid I have to swap these numbers with to number next to it. For eg. I will swap 8 to 9,16 to 17 ,32 to 33...
and print -1 if N is 2^n because in this case we cannot swap N to N+1 because we have no choice left.


So here is my working code.







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