NVIDIA HackerRank OA Frequent Problem Types Deep Dive: Sliding Window + Number Theory + GPU MCQs

NVIDIA leads the GPU and AI accelerator market, but its OA is not a "write a CUDA kernel" exotic exam — most problems land at LeetCode medium. A few role lines layer in MCQs. The catch: the hidden test cases on a NVIDIA HackerRank OA are unforgiving, and the rubric is mostly AC ratio. This guide breaks the OA into four problem types with a frequent question and Python solution per bucket.

OA at a glance

Platform     HackerRank
Time         90 min (some role lines run 60 / 120 min)
Volume       2-3 coding problems; some lines add 10 MCQs (C/C++ syntax, computer architecture)
Pass bar     >=80% test cases + at least one full AC
Scoring      Auto-grade plus human review
Re-test      Cannot re-apply to the same role within 6 months

NVIDIA OA does not rely on a CodeSignal-style score. Pass/fail is the hidden case ratio. Edge cases and large-input torture tests decide the loop.

Bucket 1 — sliding window / two pointers

Problem 1: k-th smallest in sliding window

Prompt: given nums and window size k, output the t-th smallest element per window.

from sortedcontainers import SortedList

def kth_smallest_window(nums: list[int], k: int, t: int) -> list[int]:
    sl = SortedList()
    out: list[int] = []
    for i, x in enumerate(nums):
        sl.add(x)
        if i >= k:
            sl.remove(nums[i - k])
        if i >= k - 1:
            out.append(sl[t - 1])
    return out

Complexity: O(n log k). SortedList insert / remove / index are all O(log k).

Pitfalls:

• If sortedcontainers is missing in the runtime, fall back to two heaps with lazy deletion
• Confirm whether t is 1-indexed or 0-indexed before submitting

Problem 2: subarrays with sum divisible by k

Prompt: count subarrays in nums whose sum is divisible by k.

from collections import defaultdict

def subarrays_div_by_k(nums: list[int], k: int) -> int:
    cnt = defaultdict(int)
    cnt[0] = 1
    pre = 0
    ans = 0
    for x in nums:
        pre = (pre + x) % k
        ans += cnt[pre]
        cnt[pre] += 1
    return ans

Complexity: O(n) with prefix-sum-mod-k counting.

Pitfall: negative modulo. Python's % is already normalized, but in C++ use ((pre + x) % k + k) % k.

Bucket 2 — graph / reachability

Problem: reachable cities within cost C

Prompt: directed weighted graph; sources S. Return a boolean array indicating whether each node is reachable within distance ≤ C from any source.

import heapq

def reachable_within(n: int, edges: list[tuple[int, int, int]], starts: list[int], C: int) -> list[bool]:
    g: list[list[tuple[int, int]]] = [[] for _ in range(n)]
    for u, v, w in edges:
        g[u].append((v, w))
    dist = [float("inf")] * n
    pq: list[tuple[int, int]] = []
    for s in starts:
        dist[s] = 0
        heapq.heappush(pq, (0, s))
    while pq:
        d, u = heapq.heappop(pq)
        if d > dist[u] or d > C:
            continue
        for v, w in g[u]:
            nd = d + w
            if nd <= C and nd < dist[v]:
                dist[v] = nd
                heapq.heappush(pq, (nd, v))
    return [dist[i] <= C for i in range(n)]

Complexity: O(E log V). Multi-source Dijkstra.

Key idea: model "multi-source shortest path" by seeding the heap with multiple zero-distance starts. NVIDIA loves this trick.

Bucket 3 — number theory / bit manipulation

Problem: maximum XOR subarray

Prompt: find the contiguous subarray with the maximum XOR.

class Trie:
    def __init__(self):
        self.children = {}

    def insert(self, num: int, bits: int = 32):
        node = self
        for i in range(bits - 1, -1, -1):
            b = (num >> i) & 1
            node = node.children.setdefault(b, Trie())

    def query(self, num: int, bits: int = 32) -> int:
        node = self
        ans = 0
        for i in range(bits - 1, -1, -1):
            b = (num >> i) & 1
            want = 1 - b
            if want in node.children:
                ans |= (1 << i)
                node = node.children[want]
            else:
                node = node.children[b]
        return ans

def max_xor_subarray(nums: list[int]) -> int:
    trie = Trie()
    trie.insert(0)
    pre = 0
    ans = 0
    for x in nums:
        pre ^= x
        ans = max(ans, trie.query(pre))
        trie.insert(pre)
    return ans

Complexity: O(n * 32) = O(n). 0/1 trie plus prefix XOR.

Why this matters at NVIDIA: GPU engineering frequently builds masks bit-by-bit, so bitwise problems show up regularly.

Bucket 4 — GPU / computer architecture MCQs (some role lines)

NVIDIA Drive / DGX / CUDA compiler role lines layer 10 MCQs on top of the coding section. Frequent topics:

MCQ buckets

1. C/C++ memory model: semantics of volatile / restrict / register
2. Cache hierarchy: relative latency of L1 / L2 / shared memory / global memory
3. Parallel model: CUDA warp size = 32, SIMT vs SIMD distinction
4. Compiler optimization: difference between -O2 and -O3 on loop unrolling
5. Bit operations: two's complement and IEEE 754 floating-point encoding

Sample MCQ

What is the synchronization scope of __syncthreads() in CUDA? A. The whole grid B. A single block C. A single warp D. A single thread

Answer: B (within a block). A staple of CUDA fundamentals.

Approach:

• Read the first three chapters of the CUDA C Programming Guide
• Skim 30 highly upvoted posts on the NVIDIA Developer Blog
• Map answers via OpenMP / MPI analogies if you are CUDA-light

Differences across the four role lines

Role line	Coding focus	MCQ weight
Drive / Auto	Graph traversal + state machines	10%
DGX / Cloud	Scheduling + parallelism	30%
Robotics / Edge	Geometry + sensor fusion	0%
Compiler / CUDA	Bit manipulation + number theory	50%

How OA assist plugs into NVIDIA

NVIDIA HackerRank OA is not a casual LeetCode session — hidden cases test edge boundaries question by question. Standard OA assist cadence:

1. Role-line decision: 5 minutes from the JD + recruiter notes to choose Drive / DGX / Robotics / Compiler
2. 90-min timed simulation: two real-question reps, drilling hidden-case thinking (empty arrays, max int, single element)
3. MCQ sprint: 50 questions each across CUDA / C++ / architecture, sorted by error rate
4. Runtime alignment: Python 3.10 + stdlib (no numpy / scipy) to match HackerRank, with a heapq fallback when sortedcontainers is unavailable
5. Live cue support: hidden-case candidates, bit-manipulation follow-ups, and trie templates pushed from the back channel on test day

FAQ

Q1: What is the NVIDIA OA pass bar? A: Field-observed: ≥80% of test cases plus one full AC. Below 60% almost always rejects.

Q2: Does the runtime ship sortedcontainers? A: It depends on the configuration. Try-import at the top, then fall back to heapq lazy-deletion two-heaps.

Q3: How long is review after the OA? A: 2-3 weeks typically. Drive / Robotics is faster (recruiter feedback within a week). Compiler is slower (multiple staff reviewers).

Q4: How many MCQs can you miss? A: Field-observed: keep ≥70% accuracy. For MCQ-heavy roles (Compiler / CUDA) the MCQs weigh ~30% of the total — do not concede.

Q5: Can you retake the OA? A: No. You must wait six months before reapplying to the same role. Treat the first attempt like the real thing.

Closing

The NVIDIA HackerRank OA is won by predicting hidden cases, not by raw problem volume. After every solution, sketch the three boundaries (empty / minimal / maximal) before submitting. If you are prepping for the NVIDIA OA, ping WeChat Coding0201 with your JD and the current loop stage — start with the role-line decision, then schedule OA assist timed-simulation reps.

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