【hackerrank】-Day 29: Bitwise AND--not finished!

30-bitwise-and
Objective
Welcome to the last day! Today, we’re discussing bitwise operations. Check out the Tutorial tab for learning materials and an instructional video!

Task
Given set . Find two integers, and (where ), from set such that the value of is the maximum possible and also less than a given integer, . In this case, represents the bitwise AND operator.

Input Format

The first line contains an integer, , the number of test cases.
Each of the subsequent lines defines a test case as space-separated integers, and , respectively.

Constraints

Output Format

For each test case, print the maximum possible value of on a new line.

Sample Input

3
5 2
8 5
2 2
Sample Output

1
4
0
Explanation

All possible values of and are:

The maximum possible value of that is also is , so we print on a new line.

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Current Buffer (saved locally, editable)

#!/bin/python3

import math
import os
import random
import re
import sys

def bitwiseand(n, k):
    if n <= k:
        print(0)
    else:
        s = {}
        for i in range(n):
            s[i]=i+1
        bitwise_and = []
        num = 1
        for i in range(n):
            if num >= n:
                break
            for j in s:
                bitwise_and.append(num & j)
            num += 1
            s.pop(i)
        bitwise_and_new = list(set(bitwise_and))
        for b in range(len(bitwise_and_new)):
            if bitwise_and_new[b] >= k:
                print(bitwise_and_new[b-1])
                break

if __name__ == '__main__':
    # the bad func ----Runtime Error :(
#     t = int(input())

#     for t_itr in range(t):
#         nk = input().split()
#         n = int(nk[0])
#         k = int(nk[1])
#         bitwiseand(n, k)

    # the best func, tell me why?
    T = int(input().strip())
    for _ in range(T):
        n , k = map(int , input().split())
        print(k-1 if ((k-1) | k) <= n else k-2)