This approach uses a dynamic programming (DP) solution where we maintain a 2D array called dp. The element dp[i][j] represents the maximum number of lines that can be drawn without crossing up to the i-th element of nums1 and the j-th element of nums2.

The DP formulation is based on: if nums1[i] == nums2[j], then dp[i][j] = dp[i-1][j-1] + 1. Otherwise, dp[i][j] = max(dp[i-1][j], dp[i][j-1]).

The final answer is found in dp[m][n], where m and n are the lengths of nums1 and nums2 respectively. This approach runs in O(m * n) time complexity and uses O(m * n) space.

This approach is an optimization of the basic dynamic programming solution by reducing space complexity. Instead of utilizing a 2D array, we can use a single-dimensional array to store only the current and previous row information, thus reducing the space usage. This takes advantage of the fact that the current DP state depends only on the previous state, not all preceding states.

We maintain two 1D arrays, and they get updated in each iteration over the second array, resulting in reduced space complexity to O(min(m, n)).