And Then There Were K (Codeforces 1527A)
Tags: Others, bitmasks

Given an integer $$$n$$$, find the maximum value of integer $$$k$$$ such that the following condition holds:  $$$n$$$ & ($$$n-1$$$) & ($$$n-2$$$) & ($$$n-3$$$) & ... ($$$k$$$) = $$$0$$$  where & denotes the bitwise AND operation.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 3 \cdot 10^4$$$). Then $$$t$$$ test cases follow.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^9$$$).

Output: For each test case, output a single integer — the required integer $$$k$$$.

Note: In the first testcase, the maximum value for which the continuous & operation gives 0 value, is 1.In the second testcase, the maximum value for which the continuous & operation gives 0 value, is 3. No value greater then 3, say for example 4, will give the & sum 0.   $$$5 \, \& \, 4 \neq 0$$$,  $$$5 \, \& \, 4 \, \& \, 3 = 0$$$. Hence, 3 is the answer.

Sample Inputs:
3
2
5
17

Sample Outputs:
1
3
15

Solution (Python):
import sys
input = sys.stdin.buffer.readline
for t in range(int(input())):
    N = int(input())
    print((1 << N.bit_length() - 1) - 1)

Solution (Python):
for i in [*open(0)][1:]:
    print(2 ** len(bin(int(i))) // 8 - 1)

Solution (Python):
for i in [*open(0)][1:]:
    print(2 ** len(bin(int(i))) // 8 - 1)