Microsoft

The interviewer opened with an advanced linear matroid optimization problem: "Given an array of n positive 64-bit integers and an integer k, find a subset of at most k integers whose bitwise XOR sum is non-zero, but subject to a cost constraint where every element has an associated removal penalty, maximizing the total retained weight under linear independence." I pointed out that treating this as a generic 0/1 knapsack with XOR states requires maintaining a 2^64 state space, which is computationally impossible. I reframed the task through Matroid Theory: linear independence over GF(2) forms a Vector Matroid, and restricting the subset size to at most k forms a Uniform Matroid of rank k. Because the intersection of a Vector Matroid and a Uniform Matroid with rank constraint remains a valid matroid (specifically, a truncated linear matroid), the greedy choice property holds unconditionally. The interviewer followed up: "Walk me through the exact greedy selection loop, and explain how you verify independence in real time without matrix re-inversion." I explained that we sort all candidate elements in descending order of their retention weights. We maintain a Linear Basis of at most 64 basis vectors. For each element in sorted order, we attempt to insert it into the basis by iteratively XORing it against existing pivots from bit 63 down to 0; if the reduced value does not collapse to 0, it is linearly independent, so we insert the pivot, increment our selected subset counter, and retain its weight. If it reduces to 0, adding it would introduce a linear dependency (a cycle in GF(2)), so we discard it. We break as soon as k elements are accepted or all candidates are exhausted. He verified that sorting takes O(N log N) time and the 64-bit basis insertions take O(N * 64) = O(N) operations, guaranteeing an optimal O(N log N) runtime and O(1) auxiliary space. He then shifted to a computational geometry and circular sweep challenge: "Given n circular disks on a 2D plane, each specified by a center coordinate (x, y) and radius r, compute the total area of the union of all disks without undercounting overlapping lenses or overcounting nested intersections." I noted that discrete grid sampling runs out of precision and Monte Carlo integration only yields an approximation. I proposed Green's Theorem paired with an Angular Sweep-Line on Circle Arcs. The interviewer cut in: "How does Green's theorem convert this 2D area computation into 1D boundary integrals, and how do you determine which boundary arcs are active?" I broke down the calculus and mechanics: by Green's Theorem, the area of a closed region bounded by piecewise smooth curves can be computed as the line integral (1/2) * integral(x dy - y dx) along the exterior boundary. For each circle i, we compute its geometric intersections with all other circles j. Each intersecting circle j cuts out an angular interval [theta_start, theta_end] on circle i's perimeter where circle i lies strictly inside circle j. After collecting all occluded angular intervals for circle i, we run a 1D interval union on the range [0, 2*pi] to identify the surviving, un-occluded circular arcs that form the true exterior boundary of the union. We then directly integrate the circular differential form along each active arc in closed analytic form and sum the contributions across all circles. If a circle is completely swallowed by another circle, it generates no active arcs and is skipped entirely. He had me trace the interval union on two intersecting circles, proving that with N circles generating at most O(N^2) pairwise intersections, sorting the angles on each circle bounds the entire exact area calculation to O(N^2 log N) time and O(N^2) memory. For the final challenge, he introduced an algebraic graph problem: "Given an undirected unweighted graph with n vertices, determine whether the graph contains an induced cycle of length at least 4 (a chordless cycle, or hole) in sub-quartic time." I noted that testing all subsets of vertices takes exponential time, and naively finding cycles with BFS/DFS can accidentally traverse cross-edges (chords) rather than chordless paths. I proposed an algebraic approach utilizing Matrix Multiplication and Node Neighborhood Partitions. The interviewer challenged me: "Walk me through how triad relationships and matrix powers identify whether a cycle contains chords, or show how Lexicographic BFS resolves chordal graph recognition." I explained that a graph lacks chordless cycles of length >= 4 if and only if it is a Chordal Graph. Chordal graphs possess a Perfect Elimination Ordering (PEO)—an ordering of vertices v_1, v_2, ..., v_n such that for every vertex v_i, its neighbors that appear later in the ordering form a complete clique (no missing edges). We can test for the existence of a PEO in linear O(V + E) time using Lexicographic Breadth-First Search (Lex-BFS) or Maximum Cardinality Search (MCS). In Lex-BFS, we maintain an ordered sequence of vertex sets, repeatedly selecting the vertex with the lexicographically largest label, assigning it the next spot in the ordering, and refining the remaining sets. Once the candidate ordering is produced, we verify the clique condition: for each vertex u, let v be its earliest neighbor occurring after u in the PEO; we check whether all other later neighbors of u are also adjacent to v. If this adjacency check holds for all vertices, the graph is chordal (no holes exist); if any verification fails, the discrepancy immediately certifies an induced cycle of length >= 4. He approved the structural reduction, verifying that the entire chordless cycle detection completes in strict O(V + E) time and O(V + E) space.