<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers.</p>

<p>We call a subarray of <code>nums</code> <strong>nice</strong> if the bitwise <strong>AND</strong> of every pair of elements that are in <strong>different</strong> positions in the subarray is equal to <code>0</code>.</p>

<p>Return <em>the length of the <strong>longest</strong> nice subarray</em>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p><strong>Note</strong> that subarrays of length <code>1</code> are always considered nice.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,8,48,10]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest nice subarray is [3,8,48]. This subarray satisfies the conditions:
- 3 AND 8 = 0.
- 3 AND 48 = 0.
- 8 AND 48 = 0.
It can be proven that no longer nice subarray can be obtained, so we return 3.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,5,11,13]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The length of the longest nice subarray is 1. Any subarray of length 1 can be chosen.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

from itertools import combinations
class Solution:
    # Time: O()
    # Space: O(), where n = len(nums)
    def longestNiceSubarray(self, nums: List[int]) -> int:
        
        n = len(nums)

        def and_op(l, r):
            comb = combinations(nums[l:r], 2)
            for a, b in comb:
                if a & b != 0:
                    return 0
            return 1
        # Since n <= 10^5 -> floor(log2(10^5)) + 1 = 30 bits
        # Try every possible length of nice subarray 
        for length in range(min(30, n), 1, -1):
            if and_op(0, length):
                return length
            for r in range(length, n+1):
                if and_op(r-length, r):
                    return length

        return 1
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