<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, return <em>the maximum length of a subarray that appears in <strong>both</strong> arrays</em>.</p>

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

<pre>
<strong>Input:</strong> nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The repeated subarray with maximum length is [3,2,1].
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The repeated subarray with maximum length is [0,0,0,0,0].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 100</code></li>
</ul>

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        def create_sub_arr(nums, k) -> List: 
            res = []
            for i in range(len(nums) - k + 1): 
                res.append(nums[i : i + k]) 
            return res  
        
        def check_sub_equal(list1: List, list2: List) -> bool: 
            for arr1 in list1: 
                for arr2 in list2: 
                    if arr1 == arr2: 
                        return True 
            return False 
        
        def check_k(nums1: list, nums2: list, k) -> bool: 
            sub_arrs_1 = create_sub_arr(nums1, k) 
            sub_arrs_2 = create_sub_arr(nums2, k) 
            return check_sub_equal(sub_arrs_1, sub_arrs_2)
        
        min_length = min(len(nums1), len(nums2)) 

        # Do binary search
        left = 0 
        right = min_length 
        while left <= right: 
            mid = (left + right) // 2 
            if check_k(nums1, nums2, mid): 
                left = mid + 1 
            else: 
                right = mid - 1 
        
        return right



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