Maximum length of repeated subarray

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        m = len(nums1) + 1
        n = len(nums2) + 1
        dp = [[0] * n for _ in range(m)]
        res = 0
        for i in range(1, m):
            for j in range(1, n):
                if nums1[i - 1] == nums2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                res = max(res, dp[i][j])

        return res

Maximum Length of Repeated Subarray

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Given two integer arrays nums1 and nums2, return the maximum length of a subarray that appears in both arrays.

 

Example 1:

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

Example 2:

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

 

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 100