718. Maximum Length of Repeated Subarray

Define the subproblem dp[i][j] = the length of the longest common subarray that ends exactly at nums1[i-1] and nums2[j-1]. The recurrence relation is dp[i][j] = dp[i-1][j-1] + 1 if nums1[i-1] == nums2[j-1], else dp[i][j] = 0 — the mismatch resets the chain because subarrays must be contiguous. Base cases dp[0][j] = dp[i][0] = 0. Maintain a running max as you fill the table; the answer is that max, not dp[m][n]. This is the structural difference from LCS (which carries forward through mismatches via max(dp[i-1][j], dp[i][j-1])). Time O(m * n), space O(m * n); reduce to O(min(m, n)) with rolling rows or two 1D arrays.