Microsoft

The interviewer opened with an advanced non-associative range-maintenance challenge: "Design a data structure over an array of size n that supports point updates and range queries applying a non-invertible, non-associative transformation (for example, applying $A_i = \min(A_i, v)$ over a range and answering range sum queries) in sub-quadratic time." I noted that standard Segment Trees with lazy propagation require operations that easily compose and distribute over sums, whereas range modulo or range-min updates break closed-form lazy tag accumulation. I proposed Segment Tree Beats (Historical Minimum/Maximum Tree with tag-condition pruning). The interviewer followed up: "What additional state variables must each node maintain, and what exact three-way branching condition guarantees sub-linear amortized runtime?" I explained that each node maintains the maximum value `max1`, the second strictly smaller maximum value `max2`, the frequency of the maximum value `cnt_max`, and the segment sum. When applying $A_i = \min(A_i, v)$ on a segment: if $v \ge \text{max1}$, the operation is a no-op and we return immediately; if $\text{max2} < v < \text{max1}$, the update affects only the elements equal to `max1`, so we directly adjust the sum by $(\text{max1} - v) \cdot \text{cnt\_max}$, replace `max1` with $v$, post a lazy tag, and return without descending further; if $v \le \text{max2}$, the structure cannot determine the change locally, so we push down lazy tags and recursively break into both children. Through potential function analysis, the number of distinct values across the tree strictly shrinks, bounding the amortized time to $O((n + q) \log n)$ with $O(n)$ space. He then shifted to high-dimensional spatial nearest-neighbor search: "You are given n points in a d-dimensional continuous space where d is between 5 and 10; build a data structure to support fast orthogonal range searches and arbitrary k-nearest-neighbor (k-NN) queries." I ruled out 2D Range Trees because their space complexity scales exponentially as $O(n \log^{d-1} n)$, which exhausts memory when $d \ge 5$. I proposed a k-d Tree (k-Dimensional Binary Space Partitioning Tree). The interviewer cut in: "Walk me through how the splitting planes are selected during construction to guarantee balance, and describe the geometric bounding-box pruning criterion during a k-NN search." I explained that during recursive construction, we cycle through the dimensions cyclically at depth level ($dim = depth \pmod d$) and partition points using the median coordinate along that dimension, determined in $O(\text{size})$ time via Quickselect. This ensures a balanced tree of depth $\lceil \log_2 n \rceil$ built in $O(d \cdot n \log n)$ time and $O(d \cdot n)$ space. To query the nearest neighbors to a target point $P$, we maintain a bounded max-priority-queue of the current $k$ best candidates. At each node, we compute the candidate distance along the node's splitting plane; we first recurse into the child half-space containing $P$. Crucially, we only recurse into the opposite child if the orthogonal Euclidean distance from $P$ to the dividing hyperplane is strictly less than the distance to the current $k$-th farthest candidate in our priority queue. This hyper-rectangular pruning drops the average search time to $O(2^d \log n)$. For the final challenge, he introduced an algebraic string periodicity and square-free factor problem: "Given a string s of length n, find all 'runs' (maximal periodic substrings of exponent at least 2) and locate all squares $u u$ occurring inside s in $O(n \log n)$ or $O(n)$ time." I pointed out that finding repetitions using standard sliding windows or suffix arrays naively checks all centers in $O(n^2)$ time, which TLEs when $n = 2 \cdot 10^5$. I proposed the Main-Lorentz Algorithm (Divide and Conquer with Suffix Structures) or the Runs Theorem via Lyndon Factorization. The interviewer challenged me: "Let's focus on the Divide and Conquer approach of Main-Lorentz. When splitting the string at mid into left half $u$ and right half $v$, how do you find all squares crossing the boundary without checking all sub-segments?" I explained that any square $w w$ of length $2L$ crossing the boundary must cover a segment in $u$ and a segment in $v$. We iterate over all possible half-lengths $L \in [1, n/2]$. For a fixed $L$, we place anchor checkpoints spaced exactly $L$ apart. A square of length $2L$ must contain at least one checkpoint inside its left half and one inside its right half. By computing the Longest Common Extension (LCE) forward and backward from these discrete anchor points using a precomputed LCP array over Suffix Arrays or dual polynomial rolling hashes in $O(1)$ time, the lengths of the forward match $l_1$ and backward match $l_2$ dictate whether a square exists: if $l_1 + l_2 - 1 \ge L$, a contiguous interval of valid square starting positions of length $2L$ is detected in $O(1)$. Summing $O(n / L)$ operations over all $L$ yields $\sum_{L=1}^{n/2} O(n/L) = O(n \log n)$ total checks. He verified the harmonic series summation and approved the runtime proof.