## Algorithm

Problem Name: 718. Maximum Length of Repeated Subarray

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`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
int findLength(vector& A, vector& B) {
// dp[i][j] is the maximum length of subarray ending with A[i] and B[j]
// dp[i][j] = (A[i] == B[j]) ? dp[i - 1][j - 1] + 1 : 0;
int m = A.size(), n = B.size();
vector>dp(m, vector(n));
int maxlen = 0;
for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++){
if(i == 0 || j == 0) dp[i][j] = (A[i] == B[j]);
else dp[i][j] = (A[i] == B[j]) ? dp[i - 1][j - 1] + 1 : 0;
maxlen = max(maxlen, dp[i][j]);
}
return maxlen;
}
};
``````
Copy The Code &

Input

cmd
nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]

Output

cmd
5

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int findLength(int[] nums1, int[] nums2) {
int result = 0;
int[][] dp = new int[nums1.length + 1][nums2.length + 1];
for (int i = nums1.length - 1; i >= 0; i--) {
for (int j = nums2.length - 1; j >= 0; j--) {
if (nums1[i] == nums2[j]) {
dp[i][j] = dp[i + 1][j + 1] + 1;
result = Math.max(result, dp[i][j]);
}
}
}
return result;
}
}
``````
Copy The Code &

Input

cmd
nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]

Output

cmd
5

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const findLength = function(A, B) {
let ans = 0;
let memo = [];
for (let i = 0; i < A.length + 1; i++) {
memo[i] = Array(B.length + 1).fill(0);
}
for (let i = A.length - 1; i >= 0; --i) {
for (let j = B.length - 1; j >= 0; --j) {
if (A[i] == B[j]) {
memo[i][j] = memo[i + 1][j + 1] + 1;
if (ans < memo[i][j]) ans = memo[i][j];
}
}
}
return ans;
};
``````
Copy The Code &

Input

cmd
nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]

Output

cmd
3

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def findLength(self, A, B):
A, res, sub = "X%sX" % "X".join(map(str, A)), 0, "X"
for num in B:
sub += str(num) + "X"
if sub in A: res += 1
else: sub = sub[sub[1:].index("X") + 1:]
return res
``````
Copy The Code &

Input

cmd
nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]

Output

cmd
3