<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>

class Solution:
    def longestNiceSubarray(self, nums: List[int]) -> int:
        bits = [0 for i in range(32)]
        best = 1
        cur = 0
        if len(nums) == 1:
            return 1
        l = 0
        for i in range(len(nums)):
            b = 1
            pos = 0
            good = True
            while b <= nums[i]:
                if b&nums[i] > 0:
                    if bits[pos] > 0 :
                        good = False
                    bits[pos] += 1
                b = b << 1
                pos += 1
            
            if good == True:
                cur += 1
                best = max(best, i-l +1)
            else:
                good == False
                while good == False:
                    b = 1
                    pos = 0
                    while b <= nums[l]:
                        if b&nums[l] > 0:
                            bits[pos] -= 1
                        b = b << 1
                        pos += 1
                    cur = 1
                    l += 1
                    if l == i:
                        break
                    b =1
                    pos = 0
                    good = True
                    
                    for j in range(32):
                        if  bits[j] > 1:
                            good = False
                            break
                        b = b << 1
                    
                    

        return best
            
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