<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:
#     def findLength(self, nums1: List[int], nums2: List[int]) -> int:
#         dp = [[0 for _ in range(len(nums1))] for __ in range(len(nums2))]
#         cols = len(nums1)
#         rows = len(nums2)
#         res = 0
#         for i in range(rows):
#             for j in range(cols):
#                 if nums1[j] == nums2[i]:
#                     if i > 0 and j > 0:
#                         dp[i][j] += dp[i-1][j-1] + 1
#                     else:
#                         dp[i][j] = 1
#                     res = max(dp[i][j], res)
#         return res

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        # logn - binary search over length of array + O(chek(m))
        # For each value of m, you will see if there is array of m length
        
        def check(m):
            # see if common subarray in nums1,nums2 of length m
            # Just because the constraints are upto 1000, we can form a set whihc holds all subarray
            # of nums1 and later iterate over nums2 to see if nums2 has anything that matches set
            nums1Set = set()
            for i in range(len(nums1)-m+1):
                nums1Set.add(str(nums1[i:i+m]))
            
            for i in range(len(nums2)-m+1):
                if str(nums2[i:i+m]) in nums1Set:
                    return True
            return False

        l = 0
        r = min(len(nums1),len(nums2))
        ans = -1

        while l<=r:
            m = (l+r)//2

            if check(m):
                # m is valid because there exists common subarray of length m in nums1,nums2
                # see if we can get higher value of m
                ans = m
                l = m+1
            else:
                r = m-1
        return ans
        
18100