718. Maximum Length of Repeated Subarray

1. Description

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

2. Example

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].

3. Constraints

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

4. Solutions

Dynamic Programming

n = nums1.size(), m = nums2.size()
Time complexity: O(nm)
Space complexity: O(nm)

class Solution {
public:
    int findLength(const vector<int> &nums1, const vector<int> &nums2) {
        vector<vector<int>> dp(nums1.size() + 1, vector<int>(nums2.size() + 1));
        int result = 0;
        for (int i = 0; i < nums1.size(); ++i) {
            for (int j = 0; j < nums2.size(); ++j) {
                dp[i + 1][j + 1] = nums1[i] == nums2[j] ? dp[i][j] + 1 : 0;
                result = max(result, dp[i + 1][j + 1]);
            }
        }

        return result;
    }
};