New Start day 4 : 03/10/2020 codechef october long POSAND

 question (source:https://www.codechef.com/OCT20B/problems/POSAND)

Positive AND

A permutation  of is beautiful if  is greater than 0 for every 1≤i<N . You are given an integer , and your task is to construct a beautiful permutation of length 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  if there is no suitable permutation of length , otherwise output  integers in a single line which form a beautiful permutation. If there are multiple answers output any of them.

Constraints

  • 1≤N≤105
  • The sum of N over all test cases does not exceed 106

Subtasks

  • 50 points : 1≤N,T≤9
  • 50 points : Original constraints

Sample Input:

3
4
3
5

Sample Output:

-1
1 3 2
2 3 1 5 4

My logic:

The Bitwise and operator '&' perform and operation on bits of 2 numbers so for example 2&1 =0  and 4&5 =4. we have to make longest sequence such that pi&pi+1 >0.  First, let's discuss where it's not possible. If we are given value of N as a power of 2 for example 2,4,8,16 etc then we cannot find any such number x such that x<N and x&N>0. since MSB or most significant bit increases with power of 2.
So for every N=power of 2 the answer would be -1.

But for any other value. let's say 17. 16&17>0, So we need to form a loop. whenever we reach the power of 2 say y we skip it and bring y+1 so the sequence looks like .... y-1,y+1, y and so on. 
for start 5 digits we use  2,3,1,5,4.
so sequence will start as 2,3,1,5,4,6,7,9,8,10,11,12,13,14,15,17,16 .....

My Code:

Comments

Post a Comment

Popular posts from this blog

LeetCode 3Sum

Question Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets. Brute force We can run 3 loops to check each possible triplets, we also need to eliminate the duplicates from both sides. this method will work but will work but will give time limit exceed after time constraints. To do this in given time limit we need to sort the array first and Create a loop that fixes the left most integer. This reduces the problem to 2sum for remaining array. For each loop iteration where duplicates are skipped, we need to do a left/right iteration using while loop. my code:
Read more

30. Substring with Concatenation of All Words

Problem Statement You are given a string s and an array of strings words. All the strings of words are of the same length. A concatenated substring in s is a substring that contains all the strings of any permutation of words concatenated. For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is not the concatenation of any permutation of words. Return the starting indices of all the concatenated substrings in s. You can return the answer in any order. Example 1: Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9] Explanation: Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6. The substring starting at 0 is "barfoo". It is the concaten...
Read more

Lowest common ancestor in a binary Tree

Approach Define a recursive function rec that takes three parameters: root (the current node), p (first node to find the lowest common ancestor for), and q (second node to find the lowest common ancestor for). Check if the current node root is null or if it is either p or q . If any of these conditions is true, return the current node root as the lowest common ancestor. Recursively call the rec function for the left subtree of the current node and assign the result to a variable l . Recursively call the rec function for the right subtree of the current node and assign the result to a variable r . Check if both l and r are not null. If so, it means that p and q are found on different subtrees of the current node, and the current node root is their lowest common ancestor. Return the current node root . If l is null, it means that both p and q are on the right subtree (or not present in the tree). Return r . If none of the above conditions are met, it means ...
Read more