<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, return <em>the maximum length of a subarray that appears in <strong>both</strong> arrays</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The repeated subarray with maximum length is [3,2,1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The repeated subarray with maximum length is [0,0,0,0,0].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        int n= nums1.size(), m = nums2.size();
        int maxLen = 0;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(nums1[i] == nums2[j]){
                    int cnt = 1;
                    int ii=i+1, jj= j+1;
                    while(ii<n && jj<m && nums1[ii] == nums2[jj]){
                        cnt++;
                        ii++; jj++;
                    }
                    maxLen = max(maxLen, cnt);
                }
            }
        }
        return maxLen;
    }
};
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