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12
answers
1
GATE CSE 2016 Set 1 | Question: 41
Let $Q$ denote a queue containing sixteen numbers and $S$ be an empty stack. $Head(Q)$ returns the element at the head of the queue $Q$ without removing it from $Q$. Similarly $Top(S)$ returns the element at the top of $S$ without removing ... = Pop(S); Enqueue (Q, x); end end The maximum possible number of iterations of the while loop in the algorithm is _______.
Let $Q$ denote a queue containing sixteen numbers and $S$ be an empty stack. $Head(Q)$ returns the element at the head of the queue $Q$ without removing it from $Q$. Simi...
34.5k
views
commented
Aug 29, 2017
DS
gatecse-2016-set1
data-structures
queue
difficult
numerical-answers
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–
4
answers
2
GATE CSE 2006 | Question: 13
A scheme for storing binary trees in an array $X$ is as follows. Indexing of $X$ starts at $1$ instead of $0$. the root is stored at $X[1]$. For a node stored at $X[i]$, the left child, if any, is stored in $X[2i]$ and the right child, if any, in $X[2i+1]$. To be able to store any binary tree on n vertices the minimum size of $X$ should be $\log_2 n$ $n$ $2n+1$ $2^n-1$
A scheme for storing binary trees in an array $X$ is as follows. Indexing of $X$ starts at $1$ instead of $0$. the root is stored at $X $. For a node stored at $X[i]$, th...
17.0k
views
commented
Aug 24, 2017
DS
gatecse-2006
data-structures
binary-tree
normal
+
–
3
answers
3
Sorted List
suppose there are 4 sorted lists of n/4 elements each. if we merge these list into a single sorted list of n elements, for the n=400 number of key comparisons in the worst case using an efficient algorithm is
suppose there are 4 sorted lists of n/4 elements each. if we merge these list into a single sorted list of n elements, for the n=400 number of key comparisons in the wors...
5.7k
views
asked
Aug 21, 2017
Algorithms
sorting
algorithms
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–
3
answers
4
GATE CSE 2006 | Question: 55
Consider these two functions and two statements S1 and S2 about them. int work1(int *a, int i, int j) { int x = a[i+2]; a[j] = x+1; return a[i+2] - 3; } int work2(int *a, int i, int j) { int t1 = i+2; int t2 = a[t1]; a[j] = ... CPU time) of work2 compared to work1 S1 is false and S2 is false S1 is false and S2 is true S1 is true and S2 is false S1 is true and S2 is true
Consider these two functions and two statements S1 and S2 about them. int work1(int *a, int i, int j) { int x = a[i+2]; a[j] = x+1; return a[i+2] - 3; }int work2(int *a, ...
10.3k
views
commented
Aug 17, 2017
Compiler Design
gatecse-2006
compiler-design
code-transformation
normal
code-optimization
+
–
3
answers
5
MadeEasy Workbook: CO & Architecture - Pipelining
Below is the screenshot of the question. I found that answer of the Q.10 is 7 and I unable to understand Q.11. Please correct me in Q.10 and Please explain Q.11 meaning and answer.
Below is the screenshot of the question. I found that answer of the Q.10 is 7 and I unable to understand Q.11. Please correct me in Q.10 and Please explain Q.11 meaning a...
3.6k
views
commented
Aug 13, 2017
CO and Architecture
made-easy-booklet
co-and-architecture
pipelining
+
–
1
answer
6
madeeasy workbook
if a disk system has an average seek time of 30ns and rotation rate of 360RPM. each track of the disk has 512 sectors each of size 512 Bytes. what is the time taken to read 4 successive sectors,also compute the effective data transfer rate a) 0.0843 sec , 1536 kbps b) 0.123 sec , 1436 kbps c) 0.156 sec ,1326 kbps d) 0.135 sec , 1252 kbps
if a disk system has an average seek time of 30ns and rotation rate of 360RPM. each track of the disk has 512 sectors each of size 512 Bytes. what is the time taken to re...
2.6k
views
asked
Aug 11, 2017
1
answer
7
IEEE floating point
in IEEE floating point representation, all the exponent bits are one and mantissa bits are non zero. this represent A) 0 B) infinity C) denormalized value D) error
in IEEE floating point representation, all the exponent bits are one and mantissa bits are non zero. this representA) 0B) infinityC) denormalized valueD) error
1.3k
views
asked
Aug 11, 2017
5
answers
8
GATE CSE 2000 | Question: 2.12
The following arrangement of master-slave flip flops has the initial state of $P, Q$ as $0, 1$ (respectively). After three clock cycles the output state $P, Q$ is (respectively), $1, 0$ $1, 1$ $0, 0$ $0, 1$
The following arrangement of master-slave flip flopshas the initial state of $P, Q$ as $0, 1$ (respectively). After three clock cycles the output state $P, Q$ is (respect...
11.4k
views
answered
Aug 1, 2017
Digital Logic
gatecse-2000
digital-logic
circuit-output
normal
flip-flop
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