By far the best solution here. I would be curious to know what is the tought process that brought you there ?

It will be really helpful if you could explain the idea behind this solution..

When k is ODD , k-1 is EVEN , k-1 can always be reached by (k-1) & k.

In binary form:
    k   = 11
    k-1 = 10
    k-1 == (k-1) & k

That is , ((k-1) | k) is always k. And ((k-1) | k) <= n is always TRUE.

When k is EVEN , k-1 is ODD , k-1 can only be reached if and only if ((k-1) | k) <= n is TRUE

Why?

In binary form:
    k   = 10110
    k-1 = 10101
    pos = 10111
    k-1 == (k-1) & pos

You can get k-1 if pos <= n is TRUE. And you can get pos by ((k-1) | (k-1+1)) , that is , ((k-1) | k). Otherwise , you just need to follow the process above when k is ODD (because k-1 is ODD) , then you get the answer k-2.

In brief , you can just do the test ((k-1) | k) <= n to determine the answer.

BF solution == Basic Feasible Solution

Please describe acronyms when you use them so that beginners can understand what you are saying. Thank you for the excellent solution.

I love that each challenge is opening me up to new concepts I hadn't encountered before. This was my brute force attempt but times out on 2, 3 and 4

from itertools import combinations

for a0 in range(t):
    n,k = input().strip().split(' ')
    n,k = [int(n),int(k)]
    print (max([a&b for a,b in combinations(range(1,n+1), 2) if a&b < k]))

Brilliant! So elegant :)

The whole point of this was to use bitwise AND, but I don't see any bitwise operations in your solution at all.

def getAnswer(n, k):
    power = int(log(n, 2)) - 1
    floor = 2**power

    maxNum = 0
    for i in range(floor+1, n+1):
        for j in range(1, i):
            test = i&j
            if test > maxNum and test < k:
                maxNum = test
        if maxNum == k-1:
            break
    return maxNum

This is ugly and I hate it, but it does at least match the requirements of the problem and arrives at the solution within the allotted time for all test cases.

We can use print without if:

print(k-1-int((k-1|k)>n))

Brute force solutions works here.

Your solution passes all the tests. But it would fail on k > n, for example k = 8, n = 5 must end up in 4, but in 7 it does. Moderators should add such a test. So the better solution is:

for _ in range(int(input())):
    n, k = map(int, input().split())
    print(
        (k-1 if (k-1)|k <= n else k-2)
        if k <= n
        else (n-1) & -2
    )

I spent several hours concocting my solution. It worked; howerver, whean I saw this solution I was taken aback in awe. So much beauty! Kudows on that!

Wow really impressive !

Beautiful ! Nice Brain..

While this solution works and this forum has good explanations of why it works, the results don't make sense if the value entered for k is greater than the value for n.

Of course, many will say the constraints include "2 ≤ k ≤ n". That's what we should expect in the input. So that's what the solutions are designed to work with. But what should the code do when k > n?

I think the code should check that the values are within reason and either give an error or do something appropriate. In the case of this specific problem, k is supposed to be the upper limit of the number to be found. I think it's acceptable to set k equal to n and continue the program as before.

I wonder why HackerRank doesn't include bad data in its tests. The ability to code a solution to a problem isn't an accurate representation of real world coding. Solutions should be able to handle bad data as well. I think that by not including this in challenges, new coders may not learn to write fault-tolerant code.

I tried in brute force , some testcases failed to due to timeout, your solution made my job easier.

my bruteforce solution: from itertools import combinations

def print_max_bitwise_and(n, k):
    result = 0
    max_ = 0
    for i, j in combinations(range(1, n+1),2):
        result = i&j
        if result < k and max_<result:
            max_ = result
    
    print(max_)

Could you put this into regular code or even pseudocode because I don't understand it

nice explanation. Thanks. But in no III it should be swiitching on the rightmost 0 to 1 of (k-1) and not 'SWITCHING ON RIGHTMOST 1'

For me it is the best straightforward explanation among the theads. Thank you.

An advanced understanding of binary arithmetic, I would think. Unfortunately, such cannot be imparted by coding exercises.

I did the same code in c++ but my answer isn't accepted for case 2,3,4.

hmmm. this code passes all five tests and looks nearly identical to yours?

Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for (int i = 0; i < t; i++) {
            int max = 0;
            int n = in.nextInt();
            int k = in.nextInt();
            for (int a = 1; a < n-1; a++) {
                for (int b = a+1; b <= n; b++) {
                    int val = a&b;
                    if (val > max && val < k) 
                        max = val;                    
                }    
            }
            System.out.println(max);
        }

seems happened with me as well, I've download case #4 for 5 hackos.

Your first loop can be changed to:

for(int i = 1; i < k; i++)

because A&B <= min(A,B)