# Maximum Length of Repeated Subarray

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

🙋‍♂️ Shubham Verma    🗓 July 7, 2021 ## Get the maximum length of repeated subarray

### 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

### Constraints:

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

# Solution:

``````                                    var findLength = function(A,B) {

let n=A.length;
let m=B.length

let dp = new Array(n + 1);
for (let i = 0; i <= n; i++)
{
dp[i]=new Array(m+1);
for (let j = 0; j <= m; j++)
dp[i][j] = 0;
}

for (let i = n - 1; i >= 0; i--)
{
for (let j = m - 1; j >= 0; j--)
{
if (A[i] == B[j]){
let xxx=dp[j + 1][i + 1];
if(!xxx){
xxx=0;
}
dp[j][i] = xxx + 1;
}
}
}
let maxm = 0;
for (let i = 0; i < n; i++)
{
for (let j = 0; j < m; j++)
{
maxm = Math.max(maxm, dp[i][j]);
}
}

return maxm;
};

``````

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

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