What is the brute force approach and complexity for Maximum Length of Repeated Subarray?

Try every pair of start positions and extend the run while the values match with time complexity O(N · M · min(N, M)) and space complexity O(1).

What is the optimal approach and complexity for Maximum Length of Repeated Subarray?

Keep one row; on a match row[j] = row[j − 1] + 1, on a mismatch 0; read j from high to low with optimal time complexity O(N · M) and space complexity O(M).

How do you identify the pattern for Maximum Length of Repeated Subarray?

Look for key signals in the prompt: "subarray" indicates A run of values that sit next to each other. A gap means the run starts again from 0.; "appears in both arrays" indicates The run must match in nums1 and in nums2 at the same time, so we compare pairs of positions, one from each array.; "maximum length" indicates The answer is the longest run that ends at any pair, not the value at the last pair..

Maximum Length of Repeated Subarray

Medium·LeetCode #718 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You get two arrays of integers, nums1 and nums2. Return the length of the longest subarray that appears in both arrays.

A subarray is a run of values that sit next to each other. A gap is not allowed.

Input

Two integer arrays nums1 and nums2.

Output

The length of the longest subarray that appears in both arrays.

Example 1

Input: nums1 = [1, 2, 3, 2, 1], nums2 = [3, 2, 1, 4, 7]

Output: 3

The repeated subarray is [3, 2, 1]. It sits at positions 2–4 in nums1 and at positions 0–2 in nums2.

Example 2

Input: nums1 = [0, 0, 0, 0, 0], nums2 = [0, 0, 0, 0, 0]

Output: 5

All five zeros match, so the whole array is the repeated subarray.

Constraints

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

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“subarray”

→ A run of values that sit next to each other. A gap means the run starts again from 0.

“appears in both arrays”

→ The run must match in nums1 and in nums2 at the same time, so we compare pairs of positions, one from each array.

“maximum length”

→ The answer is the longest run that ends at any pair, not the value at the last pair.

Approach Complexity Ladder

Brute Force — start at every pair, extend while equalO(N · M · min(N, M))O(1)

Try every pair of start positions and extend the run while the values match

Better — 2-D table, dp[i][j] = subarray ending at this pairO(N · M)O(N · M)

Fill an (N + 1) × (M + 1) table; a match extends the diagonal cell by one

Optimal — one row, j goes from high to lowO(N · M)O(M)

Keep one row; on a match row[j] = row[j − 1] + 1, on a mismatch 0; read j from high to low

Brute Force Intuition

Pick a start in nums1 and a start in nums2. Walk both forward together while the values match, and record how far they go. Do this for every pair of starts and keep the longest run. Each walk starts from scratch, so the same values are compared again and again.

Recognition rule for this sub-pattern:

Substring must be contiguous: on mismatch the run resets to 0.

Remember this · DP — Strings

Trigger

Longest common subsequence, edit distance, palindromes, matching with wildcards, interleaving.

Think

Look at the last characters: if they match, what do I do? If not, what are my options?

Key question

What does dp[i][j] mean, and what are the empty-string base cases?

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