Maximum Length of Repeated Subarray - Solution & Explanation

This approach uses a 2D DP array where dp[i][j] represents the length of the longest common subarray ending at nums1[i-1] and nums2[j-1]. Initialize the DP table with zeros. If nums1[i-1] == nums2[j-1], set dp[i][j] = dp[i-1][j-1] + 1, otherwise set it to 0. Track the maximum length found throughout the process.

This solution constructs a 2D array for maintaining the lengths of common subarrays at each ending position for both arrays. When matching elements are found, it extends the length from the previous subarray and potentially updates the maximum length found.

This approach involves using a sliding window over the two arrays with the help of a hashing method to check potential matches of subarrays. By varying the window size, you can find the length of the longest matching subarray without needing a full DP table.

The C solution selectively shifts and iterates over subarrays, checking integer matches directly as the two pointers overlap, recalculating lengths and ensuring a memory-efficient solution.