dp and dpPrev of size n2 + 1 with all elements set to 0.nums1 using an outer loop with index i.i, iterate over the elements of nums2 using an inner loop with index j.nums1[i - 1] == nums2[j - 1], set dp[j] to 1 + dpPrev[j - 1].nums1[i - 1] != nums2[j - 1], set dp[j] to the maximum of dp[j - 1] and dpPrev[j].dp to dpPrev.dp[n2], which represents the maximum number of uncrossed lines.