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Exams I Can Attempt Along with GATE

April 18, 2016 by ibia Leave a Comment

UGC NET: Happens in June and December but restricted to Master students (including MCA). Syllabus is bit more than GATE but difficulty level is slightly lower TIFR: For those interested in Theoretical Computer Science. Happens in December. PGEEE: IIITH is comparable to old IITs. Even if you are not financially sound you should consider taking […]

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Complexities of Basic Sorting Algorithms

April 18, 2016 by arjun Leave a Comment

Heads Up! Please don’t learn this table by heart. This is just to check your understanding and you should know how these complexities came The best case for insertion sort happens when the input array is already sorted. Each new element will then be put at the end of the array and the complexity […]

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C Coding Questions

April 15, 2016 by ibia Leave a Comment

C programming puzzles

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GATE Overflow
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Answered: GATE CSE 2014 Set 3 | Question: 39

Answered: GATE CSE 2014 Set 3 | Question: 39

I have found an alternative solution, with ref to Coreman textbook exercise, Q.12.2-8. If you can prove this (solution: https://stackoverflow.com/questions/40561235/time-complexity-of-finding-k-successors-in-bst) Then we know that since there are m nodes having value between L and H. We can also say that ‘m’ inorder successors of node with value ‘L’. So searching for L : O(log n) ‘m’ inorder successors : O(m+h) = O(m + log n) So overall time complexity : O(m + log n). a=1,b=1.c=1,d=0 Answer : 110. [...]

Answered: GATE CSE 2020 | Question: 10

https://cs.stackexchange.com/q/138479/118790 Check this out and the concept said in Peter Linz. I guess it is much harder to say that whether a LANGUAGE is $LL(k)$ or not while the question is much easier to answer if the grammar is given. See the $Example$ $7.11$ in Peter Linz, the language from the arguments given by other answers to the GATE question would make it look that the language of the grammar in Example 7.11 is not LL(k) for any $k$ but then suddenly the author comes up with a bit difficult grammar and proves that the problem with the grammar in… [...]

Answered: GATE CSE 2021 Set 2 | GA Question: 7

In this type of question, we join the missing piece, then a rectangle should be formed. In the question figure, the $L$ shape part of the shape doesn’t fit in options B and D. In option C, the size of the $U$ shape part is not the same as the question figure. So, the correct answer is $(A).$ [...]

Answered: GATE CSE 2019 | Question: 48

[youtube https://www.youtube.com/watch?v=B8ElmECvmWw]\ GATE CS 2019,Q48: Let Σ be the set of all bijections from {1, …, 5} to {1, …, 5}, where id denotes - YouTube [...]

Answered: MadeEasy Subject Test 2019: Compiler Design - Parsing

S1: False Refer this: https://stackoverflow.com/questions/34707467/is-every-ll1-grammar-also-an-lr0-grammar S2: False Refer this : https://gateoverflow.in/1251/gate-cse-2007-question-53 S3: TRUE Refer this: https://www.csd.uwo.ca/~mmorenom/CS447/Lectures/IntermediateCode.html/node3.html [...]

Answered: NIELIT 2016 DEC Scientist B (IT) - Section B: 43

D. None of the above As we have only two basic operation which are union and set difference [...]

Answered: NIELIT 2019 Feb Scientist D - Section D: 12

$\frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots+\frac{1}{20^{2}-1}$ $= \frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)} +\frac{1}{(6+1)(6-1)} +\cdots +\frac{1}{(20+1)(20-1)}$ $= \frac{2}{2}\left [\frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)}+\frac{1}{(6+1)(6-1)} +\cdots +\frac{1}{(20+1)(20-1)}\right]$ $= \frac{1}{2}\left [\frac{2}{(2+1)(2-1)} + \frac{2}{(4+1)(4-1)} +\frac{2}{(6+1)(6-1)} +\cdots +\frac{2}{(20+1)(20-1)}\right]$ $= \frac{1}{2}\left [\frac{(2+1)-(2-1)}{(2+1)(2-1)} + \frac{(4+1)-(4-1)}{(4+1)(4-1)} + \frac{(6+1)-(6-1)}{(6+1)(6-1)}+\cdots +\frac{(20+1)-(20-1)}{(20+1)(20-1)}\right]$ $= \frac{1}{2}\left [\frac{1}{(2-1)} - \frac{1}{(2+1)} + \frac{1}{(4-1)} - \frac{1}{(4+1)} + \frac{1}{(6-1)} - \frac{1}{(6+1)}+\frac{1}{(8-1)} - \frac{1}{(8+1)}+\cdots +\frac{1}{(20-1)} - \frac{1}{(20+1)}\right]$ $= \frac{1}{2}\left [\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7}+\frac{1}{7} - \frac{1}{9}+\cdots +\frac{1}{19} - \frac{1}{21}\right]$ $= \frac{1}{2}\left[\frac{1}{1} - \frac{1}{21}\right]$ $= \frac{1}{2}\left[ \frac{20}{21}\right]$ $= \frac{10}{21}$ Similar question: https://gateoverflow.in/358957/tifr2021-a-11 [...]

Answered: NIELIT 2019 Feb Scientist D - Section D: 13

$1-6+2-7+3-8+\dots\dots$ to $100$ terms $=\underbrace{1-6}_{-5}+\underbrace{2-7}_{-5}+\underbrace{3-8}_{-5}+\dots\dots \underbrace{50-55}_{-5}$ $= (1+2+\dots\dots 50) - (6+7+\dots\dots 55)$ $= (1+2+3+4+5)+\underbrace{(6+7+\dots\dots 50) - (6+7+\dots\dots 50)}_0-(51+52+53+54+55)$ $= (1+2+3+4+5)-(51+52+53+54+55) $ $=\frac{5}{2}(2(1)+(5-1)(1)) - \frac{5}{2}(2(51)+(5-1)(1))$ $(\because$ Sum of $n$ terms of A.P. sequence is $ \frac{n}{2}(2a+(n-1)d)$ $=\frac{5}{2}(6) - \frac{5}{2}(106)$ $=5(3)-5(53)$ $=15-265$ $=-250$ [...]

Answered: NIELIT 2017 OCT Scientific Assistant A - Section A: 52

Let Brian time be Y months and given Adrian invested for whole 12 months. partnership ratio 85000 ×12 : 42500 × y we need to make this as 3:1 in terms of profit so simply dividing by 85000 ×4 on both sides of ratio we'll now have 3 : y/8 that means Brian invested for 8 months [...]

Answered: NIELIT 2019 Feb Scientist C - Section D: 1

Ans is option (D) The minute covers $360^{\circ}$ in $60$ minutes. Hence, it will cover $\frac{360^{\circ}\times 35}{60}=210^{\circ}$ in $35$ minutes. Now, area of a sector with radius $r$ units and $\theta$ angle subtended at the centre ,is : $\frac{1}{2}\times r^{2}\times \theta$ units ($\theta$ in radian) . So, $210^{\circ}=210^{\circ}\times \frac{\pi}{180^{\circ}}$ radians $=\frac{11}{3}$ radians $\therefore$ Area of the sector: $\frac{1}{2}\times 100\times \frac{11}{3}=183.33$ cm$^{2}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section D: 28

Ans is option (D) Let the original price be $x$ Rs. Price after $p\%$ increase: $x(1+\frac{p}{100})$ Rs. Price after $p\%$ decrease: $x(1+\frac{p}{100})-[\frac{p}{100}\times x(1+\frac{p}{100})]=1$ (given in question) $\therefore$ $x(1+\frac{p}{100})(1-\frac{p}{100})=1$ $\Rightarrow$ $x=\frac{10000}{10000-p^{2}}$ Rs. [...]

Answered: CAT2019-2: 86

$\frac{n^{2}+7n+12}{n^{2}-n-12} = \frac{n^{2}+3n+4n+12}{n^{2}+3n-4n-12}$ $= \frac{n(n+3)+4(n+3)}{n(n+3)-4(n+3)}$ $= \frac{(n+3)(n+4)}{(n+3)(n-4)} = \frac{(n+4)}{(n-4)}$ $\therefore \frac{n^{2}+7n+12}{n^{2}-n-12} = \frac{(n+4)}{(n-4)}$ Now let’s substitute $n$ values given in options $ n = 8 $ $\frac{(n+4)}{(n-4)} = \frac{(8+4)}{(8-4)} =\frac{(12)}{(4)} = 3$ (Positive Integer) $ n = 12 $ $\frac{(n+4)}{(n-4)} = \frac{(12+4)}{(12-4)} =\frac{(16)}{(8)} = 2$ (Positive Integer) $ n = 16 $ $\frac{(n+4)}{(n-4)} = \frac{(16+4)}{(16-4)} =\frac{(20)}{(12)} \sim 1.67$ (Not an integer) $ n = 6 $ $\frac{(n+4)}{(n-4)} = \frac{(6+4)}{(6-4)} =\frac{(10)}{(2)} = 5$ (Positive Integer) Options A, B, D have produced the result as a positive integer. In question, given that we need to find “largest positive integer $n$ such that… [...]

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