Algorithms Preparation Resources for GATE CSE

July 16, 2015 by Arjun Suresh 9 Comments

Algorithms Syllabus

http://gatecse.in/wiki/Changes_to_GATE_CSE_Syllabus_in_2016#Algorithms
Searching, sorting, hashing.
Asymptotic worst case time and space complexity.
Algorithm design techniques: greedy, dynamic programming and divide‐and‐conquer.
Graph traversals, minimum spanning trees, shortest paths.

“Just a rename of Graph Traversals to Graph Search.”

Book

C.L.R.S.

Video

Shai Simonson

GO Classroom

Course Link

Recommended Tests:

GATE Overflow Test Series – GATE CSE 2022

Web Links

Previous Year Questions

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Important Questions

Important Questions in Algorithms

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2016-2 Algorithms and Pointers

GATE CSE Syllabus

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Suppose you want to transfer a file from Hyderabad to Bangalore. The size of file is 90,000 bytes. The file will be transferred in 1,500-byte data packets, for which 40 bytes are taken up with headers. The size of acknowledgement packets is 40 bytes including header. Every packet is acknowledged. The communication is bidirectional and the bandwidth is 100 Mbps (megabits/sec) in each direction. Hosts can send and receive at the same time. The latency from Hyderabad to Bangalore is 5 ms, as is the latency in the opposite direction. Assume that time to process a packet is very small,… [...]

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Given that, DIAGRAM In drum$1,$ $A$ and $B$ are in the ratio of $18 : 7.$ So, the quantity of $A = \frac{18}{25}$ The quantity of $B = \frac{7}{25}$ In final mixture, $A$ and $B$ are in the ratio of $13 : 7.$ So, the quantity of $A = \frac{13}{20}$ The quantity of $B = \frac{7}{20}$ In drum$2,$ let us assume the proportion of $B$ with respect to the overall mixture is $x.$ So, $ \frac{ \frac{7}{25} \times 3 + x \times 4}{7} = \frac{7}{20}$ $\Rightarrow \frac{21}{25} + 4x = \frac{49}{20}$ $\Rightarrow 20 (21 + 100x) = 25… [...]

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Given that, Total mixture $ = 175 \; \text{ml} + 700 \; \text{ml} $ $ = 875 \; \text{ml} $ Gopal takes out $10\%$ of the mixture and substitutes it by water of the same amount. We know that, A container contains $\text{x} \; \text{units}$ of the liquid from which $\text{y} \; \text{units}$ are taken out and replaced by water. Again from this mixture $\text{y} \; \text{units}$ are taken out and replaced by water. If this process is repeated $\text{‘n’}$ times. Then, The liquid left in the container after $\frac{ \text{n}^{th} \text{operation}}{ \text{Original quantity of the liquid in the container}}… [...]

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Let the cost price per kg of type $A,$ and $B$ be $\text{Rs.} x$ and $\text{Rs.} y$ respectively. When, $A$ and $B$ are mixed in the ratio of $3:2,$ then the profit is $10\%,$ So, selling price $:$ $ \left( \frac{3x+2y}{5} \right) \times \frac{110}{100} = 40 $ $ \Rightarrow \left( \frac{3x+2y}{5} \right) \times \frac{11}{10} = 40 \quad \longrightarrow (1) $ When, $A$ and $B$ are mixed in the ratio of $2:3,$ then the profit is $5\%.$ So, selling price $:$ $ \left( \frac{2x+3y}{5} \right) \times \frac{105}{100} = 40 $ $ \Rightarrow \left( \frac{2x+3y}{5} \right) \times \frac{21}{20} = 40 \quad \longrightarrow… [...]

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Given that, points $A$ and $B$ are $150 \; \text{km}$ apart. Each car travel at a speed of $100 \; \text{km/hr}$ for the first $50 \; \text{km}.$ $ \boxed{\text{Time} = \frac{\text{Distance}}{\text{Speed}}} $ So, $ t_{1} = \frac{50}{100} $ $ \Rightarrow \boxed{t_{1} = \frac{1}{2} \; \text{hour}} $ And, each car travel at a speed of $ 50 \; \text{kmph}$ for the $50 \; \text{km}.$ So, $ t_{2} = \frac{50}{50} $ $ \Rightarrow \boxed{t_{2} = 1 \; \text{hour}} $ And, each travel at a speed of $25 \; \text{kmph}$ for the last $50 \; \text{km}. $ So, $t_{3} = \frac{50}{25} $ $… [...]

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