Maximum Length of Repeated Subarray

Tested Python solution for LeetCode 718 with 24 pytest cases. Generate a practice environment with lcpy.

LeetCode 718, Medium. Topics: Array, Binary Search, Dynamic Programming, Sliding Window, Rolling Hash, Hash Function. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 24 parametrized pytest cases, and a playground notebook:

lcpy gen -n 718   # by problem number
lcpy gen -s maximum_length_of_repeated_subarray   # by problem name

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Problem

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

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Examples

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].

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].

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Constraints

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

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(len(nums1) * len(nums2))
    # Space: O(len(nums2))
    def find_length(self, nums1: list[int], nums2: list[int]) -> int:
        best = 0
        prev = [0] * (len(nums2) + 1)
        for a in nums1:
            cur = [0] * (len(nums2) + 1)
            for j, b in enumerate(nums2, start=1):
                if a == b:
                    cur[j] = prev[j - 1] + 1
                    if cur[j] > best:
                        best = cur[j]
            prev = cur
        return best

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Complexity

Time	Space
O(len(nums1) * len(nums2))	O(len(nums2))

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