Interview Questions Bank

Notes Lists

  • Akuna OA Coding

    view the code on nbviewer nbviewer is an open source project under the larger Project Jupyter. If you see warning message on phone, please change to Chrome browser / or view it on desktop.

  • Akuna Easy Hire

    Dishwasher Method derivative of $x^x$ Denote $y=x^x$, and take $\log$ on both sides. The derivative we need is $\frac{dy}{dx}$. And the answer is $x^x(\log{x}+1)$. Second Largest Uniform Three uniform random numbers between (0,1).

  • 2022 Aug 07

    Let $A$ and $B$ be two real $3 \times 3$ matrices such that $\operatorname{rank}(A B)=1$. Which of the following numbers cannot be a value for $\operatorname{rank}(B A)$? ANSWER: rank(BA) cannot be 3.

  • DRW OA Mixture

    Null Space of A Matrix What is the dimension of the null space of the matrix $$ \left[\begin{array}{ccccc} 1 & 3 & 3 & 4 & 5 \\ 0 & 2 & 4 & 6 & 10 \\ 2 & 6 & 6 & 8 & 10 \\ 0 & 0 & 5 & 2 & 7 \\ 0 & 3 & 6 & 9 & 15 \end{array}\right] $$ The column and row rank are always the same.

  • 2022 IMC OA

    1.Throw balls among players Four friends are playing a game of catch. Each time a player throws the ball they have an equal probability of throwing it to one of the other three players.

  • Virtu OA Coding 5 Questions

    view the code on nbviewer nbviewer is an open source project under the larger Project Jupyter. If you see warning message on phone, please change to Chrome browser / or view it on desktop.

  • 2023 IMC QT OA

    Number of Unique Sequences with Constraints There are 5 numbers, and can be used repeatedly to form a length 5 sequence. Two constraints: No number can be used more than twice No same number can be adjacent.

  • DRW 2023 QR

    Cholesky decomposition We can write matrix $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 11\end{array}\right)$ as $A=L \times L^{T}$, where $L=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ a & b & 1\end{array}\right)$.

  • Addendum_1

    def minChunks(totalPackets, uploadedChunks): if len(uploadedChunks) == 0: return bin(totalPackets).count("1") uploadedChunks.sort(key = lambda x: x[0]) remainingChunk = [] for i in range(1, len(uploadedChunks)): remainingChunk.append(uploadedChunks[i][0] - 1 - uploadedChunks[i-1][1]) if uploadedChunks[0][0] != 1: remainingChunk.