How do you solve Maximum Length of Repeated Subarray on LeetCode?

Maximum Length of Repeated Subarray (LeetCode #718) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Dynamic programming can significantly reduce the number of comparisons needed.

What is the brute force solution for Maximum Length of Repeated Subarray?

The brute force approach for Maximum Length of Repeated Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray in both arrays and compares them. This method is straightforward but inefficient as it can lead to a lot of unnecessary comparisons. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Maximum Length of Repeated Subarray?

Explain your thought process clearly when transitioning from brute force to optimal solutions. Be prepared to discuss the trade-offs between time and space complexity. Practice similar problems to recognize patterns in dynamic programming.

What are common mistakes when solving Maximum Length of Repeated Subarray?

Not initializing the dp array correctly. Forgetting to update the maximum length while filling the dp array.

#718

Maximum Length of Repeated Subarray

Medium

ArrayBinary SearchDynamic ProgrammingSliding WindowRolling HashHash FunctionDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

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Intuition

Time O(n * m)Space O(n * m)

Using dynamic programming, we can store the lengths of common subarrays in a 2D array. This allows us to build on previously computed results, significantly reducing the number of comparisons needed.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D array dp where dp[i][j] represents the length of the longest common subarray ending at nums1[i-1] and nums2[j-1].
2Step 2: Iterate through both arrays. If nums1[i-1] equals nums2[j-1], set dp[i][j] = dp[i-1][j-1] + 1.
3Step 3: Keep track of the maximum value in the dp array as you fill it.

solution.py10 lines

1def findLength(nums1, nums2):
2    m, n = len(nums1), len(nums2)
3    dp = [[0] * (n + 1) for _ in range(m + 1)]
4    max_length = 0
5    for i in range(1, m + 1):
6        for j in range(1, n + 1):
7            if nums1[i - 1] == nums2[j - 1]:
8                dp[i][j] = dp[i - 1][j - 1] + 1
9                max_length = max(max_length, dp[i][j])
10    return max_length

ℹ

Complexity note: The time complexity is O(n * m) because we fill a 2D array of size n by m, where n and m are the lengths of the two arrays. The space complexity is also O(n * m) due to the storage of the dp array.

1Dynamic programming can significantly reduce the number of comparisons needed.
2Understanding how to build on previous results is crucial in many DSA problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.