#interviewbit

InterviewBit launches Scaler Edge, a college-companion program for engg students

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In an effort to boost the availability of skilled tech talent, this program requires no prior experience in technology or coding and is open to students of all streams.

To address the gap between skills needed for high-quality technology jobs and work-readiness of graduates, Scaler (a platform by edtech startup InterviewBit) has launched what it claims is a first of its kind…

From IITH batchmates in 2010 to startup founders of InterviewBit & Scaler Academy, they’ve come a long way.

Simple Token Announces First Four Member Companies - To Launch Branded Tokens on Project’s Ecosystem #DigitalKnights

Simple Token announced the official onboarding of its first four Member Companies to launch Branded Tokens within the Simple Token ecosystem: XAIN, InterviewBit, Digital Knights and Pepo. Simple Token enables mainstream applications to deploy their own branded crypto-backed token economies, in a scalable and cryptographically-auditable manner, without needing to mint and maintain their own…

Decided to focus on more interview-type questions today.

CodeWars provides short coding challenges in the form of “kata,” and there’s a variety of languages to pick from. I’ve been working on Python, and I really don’t have a lot of experience with it so I appreciate getting the practice in. I really like this website, because the challenges so far have all been reasonable (been spending about 15-20 minutes per challenge) and they expose me to different Python libraries that I didn’t previously know about. When you sign up you also join a “clan,” i.e. a school, organization, etc. so when you submit your code for a challenge, you can see the submissions of other people in your clan for that same question. This is definitely a website I’m going to be coming back to.

<br />int** performOps(int **A, int m, int n, int *len1, int *len2) { int i, j; *len1 = m; *len2 = n; int **B = (int **)malloc((*len1) * sizeof(int *)); for (i = 0; i < *len1; i++) { B[i] = (int *)malloc((*len2) * sizeof(int)); } for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { B[i][n - 1 - j] = A[i][j]; } } return B; }

Let’s say m=3, n=4, and A : [[1,2,3,4], [5,6,7,8], [9,10, 11, 12]]

Wha…