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Suppose we have an array of numbers called nums. We have to check whether there exists any subset of the nums whose bitwise AND is a power of two or not.

So, if the input is like nums = [22, 25, 9], then the output will be True, as a subset {22, 9} the binary form is {10110, 1001} the AND of these two is 10000 = 16 which is power of 2.

To solve this, we will follow these steps −

- MAX := 32 considering there are 32 bits numbers at max
- Define a function solve() . This will take nums
- if size of nums is 1, then
- return true when nums[0] is power of 2, otherwise false

- total := 0
- for i in range 0 to MAX - 1, do
- total := total OR 2^i

- for i in range 0 to MAX - 1, do
- ret := total
- for j in range 0 to size of nums, do
- if nums[j] AND (2^i) is non-zero, then
- ret := ret AND nums[j]

- if nums[j] AND (2^i) is non-zero, then
- if ret is power of 2, then
- return True

- return False

Let us see the following implementation to get better understanding −

MAX = 32 def is_2s_pow(v): return v and (v & (v - 1)) == 0 def solve(nums): if len(nums) == 1: return is_2s_pow(nums[0]) total = 0 for i in range(0, MAX): total = total | (1 << i) for i in range(0, MAX): ret = total for j in range(0, len(nums)): if nums[j] & (1 << i): ret = ret & nums[j] if is_2s_pow(ret): return True return False nums = [22, 25, 9] print(solve(nums))

[22, 25, 9]

True

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