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# Maximum Length of Repeated Subarray

## Description

Given two integer arrays `A` and `B`, return the maximum length of an subarray that appears in both arrays.

**Example 1:**

```
Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation: 
The repeated subarray with maximum length is [3, 2, 1].
```

**Note:**

1. 1 <= len(A), len(B) <= 1000
2. 0 <= A\[i], B\[i] < 100

## Solutions

### My solution

```cpp
class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int n = A.size();
        int m = B.size();
        if(n == 0 || m == 0) return 0;
        // dp[i][j]: length of the longest common subarray of A[0..i] and B[0..j] ending with A[i] and B[j]
        vector<vector<int>> dp(n, vector<int>(m));
        int len = 0;
        for(int j = 0; j < m; ++j){
            if(A[0] == B[j]){
                len = dp[0][j] = 1;
            }
        }
        for(int i = 0; i < n; ++i){
            if(A[i] == B[0]){
                len = dp[i][0] = 1;
            }
        }
        for(int i = 1; i < n; ++i){
            for(int j = 1; j < m; ++j){
                if(A[i] == B[j]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    len = max(len, dp[i][j]);
                }
            }
        }
        return len;
    }
};
```

## More concise implementation

```cpp
class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int n = A.size();
        int m = B.size();
        // dp[i][j]: length of longest common subarray of A[0..i-1] and B[0..j-1] ending with A[i-1] and B[j-1]
        vector<vector<int>> dp(n + 1, vector<int>(m + 1));
        int len = 0;
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                if(A[i - 1] == B[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    len = max(len, dp[i][j]);
                }
            }
        }
        return len;
    }
};
```