<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:
        '''
        brute force: compare every element with every other element and keep checking sequence length 
            o(N*M^2)

        start checking sequence
            
        '''
        maxlen = 0
        maxArr = []
        if len(nums1) == 0 or len(nums2) == 0:
            return []
        i = 0 
        checked = [[0 for i in range(len(nums2))] for i in range(len(nums1))]
        while i < len(nums1):
            j = 0 
            while j < len(nums2):
                if checked[i][j]:
                    j += 1
                    continue
                count = 0
                iter_i = i
                iter_j = j
                while iter_i < len(nums1) and iter_j < len(nums2) and nums1[iter_i] == nums2[iter_j] :
                    count += 1
                    checked[iter_i][iter_j] = 1
                    iter_i += 1
                    iter_j += 1
                if count > maxlen:
                    maxArr = nums1[i:iter_i]
                    maxlen = count
                j += 1
            i += 1

        return len(maxArr)



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