# Algorithms – GATE CSE Previous Year Concerns

Fixing GATE Previous Year’s Concerns (PYQs) not just clears the ideas however likewise assists to get versatility, speed, precision, and understanding of the level of concerns usually asked in eviction test, which ultimately assists you to get excellent marks in the evaluation. Previous Year Concerns assist a prospect practice and modify for GATE, which assists fracture GATE with a great rating.

Algorithms Previous Year GATE Concerns aid in evaluating the concern pattern of a subject and marking plan in addition to it assists in time management which general boosts ball game in eviction test. With routine practice of PYQs, prospects can quickly split GATE with a great GATE Rating.

Prior to 2006, concerns asked in GATE were generally theoretical, however recently, the concerns asked were multiple-choice concerns with a single proper choice or numerous proper choices. We are wanting to offer the multiple-choice concerns that are asked in GATE.

## Algorithms GATE Previous Year Concerns

In this post, we are generally concentrating on the Algorithms GATE Questions that are asked in Previous Years with their options, and where a description is needed, we have actually likewise supplied the factor.

In Algorithms, we will handle the following ideas. We have actually likewise supplied GATE Previous Year’s Concerns on these subjects. Here is the list of those subjects in addition to their links.

Likewise, here we are going to talk about some fundamental PYQs associated to Algorithms

1. Which among the following declarations holds true for all favorable functions f( n)? [GATE CSE 2022]

( A) f( n 2) = Î¸( f( n) 2), when f( n) is a polynomial

( B) f( n 2) = o( f( n) 2)

( C) f( n 2) = O( f( n) 2), when f( n) is a rapid function

( D) f( n 2) = Î©( f( n) 2)

Service: Right response is ( A)

For more, describe GATE|CS 2022|Concern 11

2. For criteria a and b, both of which are Ï( 1 ), T( n) = T( n 1/a) + 1, and T( b) = 1. Then T( n) is [GATE 2020]

( A) Î¸( log a log b n)

( B) Î¸( log ab n)

( C) Î¸( log b log a n)

( D) Î¸( log 2 log 2 n)

Service: Right response is ( A)

For more, describe GATE|GATE CS 2020|Concern 12

3. The Floyd-Warshall algorithm for all-pair quickest courses calculation is based upon [GATE CSE 2016]

( A) Greedy Algorithm

( D) neither Greedy nor Divide-and-Conquer nor Dynamic Shows Paradigm

Service: Right response is ( C)

For more, describe GATE|GATE-CS-2016 (Set 2)|Concern 24

4. Which among the following is the reoccurrence formula for the worst-case time intricacy of the Quicksort algorithm for sorting (n â¥ 2) numbers? In the reoccurrence formulas given up the choices listed below, c is a continuous. [GATE CSE 2015]

( A) T( n) = 2T( n/2) + cn

( B) T( n) = T( n-1) + T( 0) + cn

( C) T( n) = 2T( n-1) + cn

( D) T( n) = T( n/2) + cn

Service: Right response is ( B)

For more, describe GATE|GATE-CS-2015 (Set 1)|Concern 12

5. An unordered list includes n unique aspects. The variety of contrasts to discover an aspect in this list that is neither optimum nor minimum is [GATE CSE 2015]

( A) Î¸( n log n)

( B) Î¸( n)

( C) Î¸( log n)

( D) Î¸( 1 )

Service: Right response is ( D)

For more, describe GATE|GATE-CS-2015 (Set 2)|Concern 65

6. Think about the following selection of aspects: ( 89,19,50,17,12,15,2,5,7,11,6,9,100). The minimum variety of interchanges required to transform it into a max-heap is [GATE CSE 2015]

( A) 4

( B) 5

( C) 2

( D) 3

Service: Right response is ( D)

For more, describe GATE|GATE-CS-2015 (Set 3)|Concern 65

7. The tightest lower bound on the variety of contrasts, in the worst case, for comparison-based sorting is of the order of [GATE CSE 2004]

( A) n

( B) n 2

( C) n log n

( D) n log 2 n

Service: Right response is ( C)

For more, describe Algorithms|Arranging|Concern 13

8. The issues 3-SAT and 2-SAT are [GATE CSE 2004]

( A) both in P

( B) both NP-Complete

( C) NP-Complete and in P respectively

( D) Undecidable and NP-Complete respectively

Service: Right response is ( C)

For more, describe GATE-CS-2004|Concern 30

9. An arranging strategy is called steady if: [GATE CSE 1999]

( A) It takes O( n log n) time

( B) It preserves the relative order of event of non-distinct aspects.

( C) It utilizes divide and dominates paradigm

( D) It takes O( n) area

Service: Right response is ( B)

For more, describe GATE|GATE CS 1999|Concern 12

10. For combining 2 arranged lists of sizes m and n into an arranged list of size m+ n, we need contrasts of [GATE CSE 1995]

( A) O( m)

( B) O( n)

( C) O( m+ n)

( D) O( log m + log n)

Service: Right response is ( C)

## GATE CSE Previous Year Concern Documents

These previous year’s concerns assist you in comprehending the concern patterns followed by GATE that straight assist a prospect in scoring excellent marks in GATE. Below are the discussed links of year-wise GATE Previous Concern Documents.