The DP. Only the previous row can interact with the current one, so the state is (row, mask of the previous row). With cols <= 8cols <= 8 there are 256 masks, and rows * 256 * 256rows * 256 * 256 is about 500,000 operations.

The three legality tests, each one shift:

Why this is a flow problem underneath. The conflict graph — one node per usable seat, an edge for every copying pair — is bipartite: colour each seat by the parity of its column. Every conflict, horizontal or diagonal, joins columns cc and c ± 1c ± 1, so it always joins opposite parities. There is never an odd cycle.

That matters because of König’s theorem: in a bipartite graph, the maximum independent set has size V - (maximum matching)V - (maximum matching). Seating students with no conflicts is exactly a maximum independent set on that graph. So the answer is also (number of usable seats) minus (maximum bipartite matching), and maximum bipartite matching is max flow with unit capacities — Hopcroft-Karp, or plain Hungarian augmenting paths.

Both routes are correct. The DP is $O(rows \cdot 4^{cols})$, so it lives or dies by the column count. The flow route is $O(E\sqrt{V})$ in the seat count and does not care how wide the room is. LeetCode caps the grid at 8 x 88 x 8, which is a clear vote for the DP — but naming König’s theorem is what turns a passing answer into a strong one.

• The diagonal rule is not symmetric with the vertical one. Directly in front is explicitly allowed, so there is no prev & curprev & cur test. Adding one is the most common wrong answer.
• A fully broken row contributes only the mask 0, and the DP continues through it correctly — bestbest never becomes empty because cur = 0cur = 0 always survives.
• [["#"]][["#"]] gives 0. No usable seats, and max(best.values())max(best.values()) is 0 rather than an error, because the mask-0 entry is always there.
• cur & ~freecur & ~free relies on Python’s arbitrary-precision two’s complement. It works, but cur & free != curcur & free != cur says the same thing more plainly.
• Only the previous row matters, not all rows above. Convincing yourself of that is what justifies the state.

Follow-ups you should expect: “What if there were 30 columns?” — $4^{30}$ is hopeless, so switch to the matching formulation. “What if the vertical neighbour also conflicted?” — add prev & curprev & cur to the test; the graph stays bipartite by column parity. “Knight-move conflicts?” — still bipartite by colour, so matching still applies. “Return the actual seating?” — store the predecessor mask per state and walk back. “Maximum independent set on a general graph?” — NP-hard; the bipartite structure here is the entire reason it is tractable.

The assignment problem: pair up two sets of nn items at minimum total cost. Brute force is $n!$ — about 87 billion at n = 14n = 14. The bitmask DP is $O(2^n \cdot n)$, roughly 230,000 states-times-transitions. That gap is why the constraint is 14 and not 20.

The state trick. ii and maskmask are redundant: you have placed exactly popcount(mask)popcount(mask) entries of nums1nums1, so i == popcount(mask)i == popcount(mask) always. Keeping both is clearer and memoizing on the pair is harmless — but mentioning that one number suffices shows you have understood the structure, and it halves the memo key.

The flow view. This is min-cost bipartite perfect matching. Build a source into every nums1nums1 node with capacity 1, an edge i -> ji -> j with capacity 1 and cost nums1[i] ^ nums2[j]nums1[i] ^ nums2[j], and every nums2nums2 node into a sink with capacity 1. Min-cost max-flow gives the answer, as does the Hungarian algorithm in $O(n^3)$ — polynomial, and therefore the right answer if nn were 200 instead of 14. Naming it is worth real credit.

Time $O(2^n \cdot n)$. Space $O(2^n)$.

• Greedy fails. Pairing each nums1[i]nums1[i] with whichever nums2[j]nums2[j] minimises its own XOR is wrong: [1,2][1,2] with [2,3][2,3] greedily takes 1^3 = 21^3 = 2 and then is forced into 2^2 = 02^2 = 0, which happens to be optimal, but [1,0,3][1,0,3] with [5,3,4][5,3,4] breaks it. A local choice steals the partner another element needed more.
• XOR is not monotonic, so sorting both arrays is not a valid strategy either — and that is the intuition most people reach for first.
• n = 1n = 1 returns nums1[0] ^ nums2[0]nums1[0] ^ nums2[0], which is 0 for [0][0] and [0][0].
• [1,2,3][1,2,3] with [3,2,1][3,2,1] returns 0 — a perfect reversal pairs each with its equal.
• cache_clear()cache_clear() matters because the cache is created per call here but the closure captures nums1nums1 and nums2nums2; clearing it keeps memory flat when the grader runs many cases.

Follow-ups you should expect: ”n = 200n = 200?” — the Hungarian algorithm, $O(n^3)$. “Which pairing?” — store the chosen jj per state and walk forward. “Maximise instead?” — swap minmin for maxmax; nothing else changes. “Unequal array lengths?” — pad with zero-cost dummy nodes. “Minimise the maximum pair XOR instead of the sum?” — binary search the answer and test for a perfect matching using only edges below the threshold; that is bottleneck assignment, and a genuinely different algorithm.

The word to notice is at least one. This is an edge cover, not a matching: a point may take several edges, and the two groups need not be the same size.

Two facts make the DP small, and both are worth stating as claims you can defend:

1. One edge per group-1 point is enough while sweeping. Giving a group-1 point a second edge only helps by covering some group-2 point — and the final patch-up step will cover that point anyway, at a cost no higher than any specific choice here. So the sweep never needs to branch on multiple edges.
2. The leftovers are independent. Once group 1 is exhausted, each uncovered group-2 point may attach to whichever group-1 point it likes, with no interaction between them. So each simply takes its column minimum, precomputed once.

Together the state is (group-1 index, covered mask), giving $m \cdot 2^n$ states with $n$ transitions each.

The flow view. Minimum-cost edge cover on a bipartite graph is solvable in polynomial time — it reduces to minimum-cost perfect matching on a transformed graph, and hence to min-cost max-flow. That is the answer for large inputs. With m, n <= 12m, n <= 12, $12 \cdot 4096 \cdot 12$ is about 590,000 operations, so the bitmask DP wins on simplicity.

Time $O(m \cdot 2^n \cdot n)$. Space $O(m \cdot 2^n)$.

• Greedy fails. Giving every point its own cheapest edge double-counts and overshoots: on [[1,3,5],[4,1,1],[1,5,3]][[1,3,5],[4,1,1],[1,5,3]] the row minima sum to 3 but the answer is 4, because those choices leave a group-2 point uncovered.
• The base case must be a sum, not zero. Forgetting to charge the uncovered group-2 points is the main wrong answer, and it silently under-reports.
• m <= nm <= n is promised by the problem but the code never relies on it — the sweep is over group 1 and the mask over group 2 either way.
• 1 x 11 x 1 returns the single cost, since both points need each other.
• All costs are positive, which is what makes “cover with as few edges as possible” the right instinct. With zero or negative costs the argument for one edge per point would need re-examining.

Follow-ups you should expect: ”n = 30n = 30?” — $2^{30}$ states is too many; switch to min-cost flow. “Each point needs at least kk connections?” — the mask must count, not just flag, so the state becomes base-(k+1)(k+1) digits and grows fast. “Maximise total cost instead?” — the one-edge argument breaks, because extra edges now always help; you would take every edge. “Return the connections?” — store the chosen jj per state, then add the patch-up edges. “Why not just matching?” — a perfect matching does not exist when m != nm != n, which is exactly why this is a cover.