<p>You are given an integer array <code>nums</code> of size <code>n</code>.</p>

<p>Consider a <strong>non-empty</strong> subarray from <code>nums</code> that has the <strong>maximum</strong> possible <strong>bitwise AND</strong>.</p>

<ul>
	<li>In other words, let <code>k</code> be the maximum value of the bitwise AND of <strong>any</strong> subarray of <code>nums</code>. Then, only subarrays with a bitwise AND equal to <code>k</code> should be considered.</li>
</ul>

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

<p>The bitwise AND of an array is the bitwise AND of all the numbers in it.</p>

<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,3,2,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.
</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>6</sup></code></li>
</ul>

class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        n=len(nums)
        maxLen, xMax, Len=0, 0, 0
        i=0
        while i<n:
            while i<n and nums[i]==xMax:
                i+=1
                Len+=1
            if i==n or nums[i]<xMax:
                maxLen=max(maxLen, Len)
                Len=0
            else:
                xMax=nums[i]
                maxLen=Len=1
            i+=1
        return maxLen
        
427