<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>

import math

class Solution:
    def longestNiceSubarray(self, nums: List[int]) -> int:
        maxBits = 32
        #print("Start:", bin((2**(maxBits+1)-1)))
        available = (2**(maxBits+1)-1) & (~nums[0])
        #print("Adding:", bin(nums[0]), "Got:" , bin(available))
        maxLen = 1
        ans = 1
        for i in range(1, len(nums)):
            #print("Adding:", bin(nums[i]))
            if (nums[i] | available) == available:
                maxLen += 1
                available = available & (~nums[i])
            else:
                while not ((nums[i] | available) == available):
                    if maxLen < 0:
                        print("Security break")
                        break
                    #print("Removing:", bin(nums[i-maxLen]))
                    last = nums[i-maxLen]
                    available = available | last
                    maxLen -= 1
                maxLen += 1
                available = available & (~nums[i])
            #print("Got:", bin(available), "maxLen:", maxLen)
            ans = max(ans, maxLen)
        return ans
553