<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:
        n = len(nums)
        left = 0
        maxLen = 1
        lastIndices = {}
        curr_or = 0
        for i in range(n):
            # window expansion
            curr_or |= nums[i]

            # window shrinkage
            pos = 0
            curr = nums[i]
            while curr:
                if curr & 1 == 1: # current position has one in binary representation
                    if pos in lastIndices and lastIndices[pos] >= left: ## left boundary of window is left to lastindex of 1 at current position
                        left = lastIndices[pos] + 1
                    lastIndices[pos] = i
                curr >>= 1
                pos += 1

            maxLen = max(maxLen, i-left+1)

        return maxLen


        # n = len(nums)
        # left = 0
        # maxLen = 1
        # p1 = 0
        # p2 = 1
        # while p2 <= n-1:
        #     while p2 <= n-1 and nums[p2]&nums[p1] == 0:
        #         p2 += 1

        #     print(p1)
        #     print(p2)

        #     maxLen = max(maxLen, p2-p1)
        #     if p2 == n:
        #         return maxLen
            
        #     if p1 == p2-1:
        #         p2 += 1
        #         p1 += 1
        #     else:
        #         p1 = p2
        #         p2 += 1

        # return maxLen




        
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