<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>

def findLength_ij(nums1, nums2, i, j):
    same = (nums1[i] == nums2[j])
    k = 0
    while same:
        i += 1
        j += 1
        if (i < len(nums1)) and (j < len(nums2)):
            same = (nums1[i] == nums2[j])
        else:
            same = False
        k += 1
    return k

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        n, m = len(nums1), len(nums2)
        mem = [[-1]* m for _ in range(n)]

        l1, r1 = 0, 0
        l2, r2 = 0, 0
        maxN = 0
        for i in range(n):
            for j in range(m):
                if mem[i][j] == -1:
                    length_ij = findLength_ij(nums1, nums2, i, j)
                    for k in range(length_ij):
                        # print(i,j,length_ij)
                        mem[i+k][j+k] = length_ij
                    maxN = max(maxN, mem[i][j])
        return maxN
        
24400