LeetCode #718 — MEDIUM

Maximum Length of Repeated Subarray

Move from brute-force thinking to an efficient approach using array strategy.

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The Problem

Problem Statement

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

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming · Sliding Window

Example 1

[1,2,3,2,1]
[3,2,1,4,7]

Example 2

[0,0,0,0,0]
[0,0,0,0,0]

Related Problems

Minimum Size Subarray Sum (minimum-size-subarray-sum)
Longest Common Subpath (longest-common-subpath)
Find the Maximum Length of a Good Subsequence II (find-the-maximum-length-of-a-good-subsequence-ii)
Find the Maximum Length of a Good Subsequence I (find-the-maximum-length-of-a-good-subsequence-i)

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming. dp[i][j] will be the longest common prefix of A[i:] and B[j:].
The answer is max(dp[i][j]) over all i, j.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #718: Maximum Length of Repeated Subarray
class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        int[][] f = new int[m + 1][n + 1];
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (nums1[i - 1] == nums2[j - 1]) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                    ans = Math.max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #718: Maximum Length of Repeated Subarray
func findLength(nums1 []int, nums2 []int) (ans int) {
	m, n := len(nums1), len(nums2)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if nums1[i-1] == nums2[j-1] {
				f[i][j] = f[i-1][j-1] + 1
				if ans < f[i][j] {
					ans = f[i][j]
				}
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #718: Maximum Length of Repeated Subarray
class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        ans = 0
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if nums1[i - 1] == nums2[j - 1]:
                    f[i][j] = f[i - 1][j - 1] + 1
                    ans = max(ans, f[i][j])
        return ans

// Accepted solution for LeetCode #718: Maximum Length of Repeated Subarray
struct Solution;

impl Solution {
    fn find_length(a: Vec<i32>, b: Vec<i32>) -> i32 {
        let n = a.len();
        let m = b.len();
        let mut dp: Vec<Vec<i32>> = vec![vec![0; m + 1]; n + 1];
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                if a[i] == b[j] {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                    res = res.max(dp[i + 1][j + 1]);
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let a = vec![1, 2, 3, 2, 1];
    let b = vec![3, 2, 1, 4, 7];
    let res = 3;
    assert_eq!(Solution::find_length(a, b), res);
}

// Accepted solution for LeetCode #718: Maximum Length of Repeated Subarray
function findLength(nums1: number[], nums2: number[]): number {
    const m = nums1.length;
    const n = nums2.length;
    const f = Array.from({ length: m + 1 }, _ => new Array(n + 1).fill(0));
    let ans = 0;
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            if (nums1[i - 1] == nums2[j - 1]) {
                f[i][j] = f[i - 1][j - 1] + 1;
                ans = Math.max(ans, f[i][j]);
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.