<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:
        LENGTH = len(nums)
        # MAKE A FUNCTION THAT CHECKS FOR OVERLAPPING SET BITS
        def overlapping_bits(n1 , n2):
            return  (n1 & n2)
        # You are given an array nums consisting of positive integers.
        # ex : 1 3 8 48 10
        # bin: 1 11 1000 110000 1010
        # Ans: n 11 1000 110000  n
        # ININTIALIZE SLIDING WINDOW (ARRAY LIMITS) AT ZERO
        l , r = 0 , 0
        # INITIALIZE XORSUM TO THE FIRST ELEMENT
        xor_of_window = 0
        # INITIALIZE MAXWINDOW TO 1
        max_window = 1

        # ALL SET BITS MUST BE IN DIFFERENT POSITIONS
        # LOOP UNTIL THE RIGHT LIMIT REACHES THE END OF THE ARRAY
        while r < LENGTH:
            # REMOVE THE LEFT LIMIT OF THE ARRAY UNTIL IT BECOMES VALID
            while overlapping_bits(xor_of_window , nums[r]):
                xor_of_window = xor_of_window^nums[l]
                print(bin(xor_of_window), 'inside loop')
                l += 1
            # TAKE THE NUMBER
            xor_of_window ^= nums[r]            

            r += 1
            # UPDATE MAXWINDOW
            max_window = max(max_window , r-l)
        # RETURN MAXWINDOW
        return max_window

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