define the time and space complexity of an algorithm and about what can i do when the time and space complexity are given and i should write the code corresponding to the restrictions , which i find very difficult.
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What would the worst, average and best case space complexity be for a data structure of type map<string, vector<int> > in big O notation? I'm parsing through a document and storing each word as a key and im attaching an associated int (in a vector) to it as the value.
View 4 Replies View Relatedthis function has to check if an array is not descending it has to be recursive and it's space complexity has to be O(log(n))
Code:
int is_sorted(int a[ ], int n) {
int check=n-1;
if(n==1 || n==0) return 1;
if(a[check]<a[check-1])return 0;
else return(is_sorted(a,n-1));
}
this is what i came up with. i know it's space complexity is O(n) how do i make it O(log(n))?
I was just going through Radix Sort algorithm. Though I got the logic and the concept used in it, I am still pretty much confused about the space complexity being O(k+n) for this algorithm. (where, k is the max no of digits in a number, and n is the no. of inputs to be sorted).
View 1 Replies View Relatedwould like to know that my algorithm has asymptotic complexity recursive?
int solve(pair<int,int> actual)
{
if(actual.first == PuntoInicio.first && actual.second == PuntoInicio.second)
return 1;
[Code]....
I have a very simple program the time complexity of the function that I used in this program is O(mn)because it has a nested loop, I really need to reduce the time complexity to O(n)
[code=c++]
#include <iostream.h>
#include<stdlib.h>
int *char_count( const char* DNA, const int *starts, const int *ends, char letter);
int main()
[Code].....
We were discussing how to find average time complexity of different algorithms. Suppose I've a the following nested loop
for(int i=0;i<n;i++)
{
min = i;
[Code].....
Now the outer loop will iterate N times. the inner loop will always iterate 3 times no matter what the value of N is. so the worst case time complexity should be O(n^2) but what about the average case and the best case? I mean in terms of searching we can figure out and distinguish between the worst,best and average as it depends upon the number of comparisons. But here how can the average and best case be different then the worst case.
Now I know there's a simple way to calculate highest value. But here i have this algorithm which i understood part of it.
I know that *largest is point to the first element of begin which is 5. so in the first if statement its 5<5 at the first iteration ?
// i marked the parts i didnt understand with " //"
Code:
#include <stdio.h>
int *print(int *begin ,int *end ){
if(begin==end)
return 0;
int *largest = begin;
[Code] ....
I have an school assignment that asks me to measure the most famous sorting algorithms for performance in terms of number of steps and CPU running time. ( Here I'm testing for running time)
I decided to test for bubble sort first:
#include <iostream>
#include <ctime>
using namespace std;
void bubbleSort(int ar[], int size) {
int temp;
[Code] ....
So basically what I want to know is:
1. Is this clock function giving the correct CPU running time?
2. Is there any way to write code that would measure the number of steps for each algorithm?
3.I need to test it for number of integers=100 then 200, then 300... Well you get my point, and I don't want to have to actually input 200 numbers with my keyboard. Is there any way to generate as many entries as I want?
i writing a code which will calculate the difference between time_in and time out. Can i use a 2 dimensional array for example
int time_in[hours][minutes]
int time_out[hours[minutes]
I want to calculate processing time in milliseconds as most of function I found give me in second. I am new user to C++, the code will be running under linux.
View 5 Replies View RelatedI'm challenged to write a program which calculates the time difference between two daytime.
compiler doesn't give an error, yet the warning: "format '%d' expects argument of type 'int *', but argument 3 has type 'int' [- Wformat]" for lines 24 and 26
However, by entering the first time the program crashes anyway. so I assume I do really need some pointer to make it read from the console can you see where my problem is?
Code:
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int Second;
int Minute;
int Hour;
}Time;
[Code]...
I have been trying for several months to create a simple SNTP single Client/Server based on RFC5905. Finally I manage to make it work at least I think it works correctly, but when I tried to test my code against a real NTP server (e.g. 0.se.pool.ntp.org:123) the timestamps that I am receiving need to be recalculated. I have tried several different approaches but no matter for 3 days now but no matter what I tried nothing yet.
The problem that I am having in not so much programming error is more about knowing how the NTP server operates.
How to convert the NTP timestamp to Unix epoch timestamp?
Syntax to execute the Server e.g. ./server 127.0.0.1:5000 and Client e.g. ./client 127.0.0.1:5000
Syntax to execute the Client against a real NTP server e.g. ./client 0.se.pool.ntp.org:123
Sample of working code Client:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <errno.h>
#include <string.h>
#include <sys/types.h>
[code]....
Sample of printed output when I use Server and Client.
Reference Identifier 0 LOCL
Reference Timestamp 0.0 1426637081.3564398733
Originate Timestamp 1426637087.3570333925 1426637087.3570333925
Receive Timestamp 1426637087.3570334078 1426637087.3570334003
Transmit Timestamp 1426637087.3570333925 1426637087.3570334046
Sample of printed output when I am probing a real NTP server (e.g. 0.se.pool.ntp.org:123).
Reference Identifier 0 $
Reference Timestamp 0.0 3879449560.3503094062
Originate Timestamp 1426637090.3573978972 1426637090.3573978972
Receive Timestamp 1426637090.3573992772 2722083800.781009125
Transmit Timestamp 1426637090.3573978972 2722083800.937312997
It is supposed to calculate and graph time vs. distance. The problem I have is graphing, if I put too small values for Vi, which is the initial velocity, it just graphs one or two points, because t is too small. I already tried to adjust t with 'if(t<10) t=t*10; if (t>100) t=t/10;', But it seems not to solve my problem.
Other truble I have is that, as the graph is parabola, core values are very close together, and when plotted, these values are in the same line on the screen.
Here is my code:
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main() {
float x, y, hmax, t, amax, vi, v, a, b, g, ang; /*x= x coordinate, y= y coordinate, hmax= maximum height,
[Code] .....
I'm trying to create a program to calculate the flight time and fuel burn of an airplane. However when I repeated the whole calculation, it did not show any data or result like the previous calculation. In addition, the "If" statement that I used to test the "fuelRemain" doesn't work properly. I can't figure out why.
Here are the outputs [URL]
#include <iostream>
#include <fstream>
#include <string>
[Code].....
Im just wondering how would i get a formula to calculate the total size and proportion of how far the media player has been played, in proportion with the size of the sizeable form, i need an int for the:
Width of formLength of movieHow long has been played by user
I have this code so far:
//The current position played! - Within the timer.tick event arg!
string splayed = axWindowsMediaPlayer1.Ctlcontrols.currentPosition.ToString().Split('.')[0];
int iplayed = Convert.ToInt32(splayed + 1);
//The total time of the movie/audio.
string stotal = axWindowsMediaPlayer1.Ctlcontrols.currentItem.duration.ToString().Split('.')[0];
int itotal = Convert.ToInt32(splayed);
[Code] ....
I have the "AxWindowsMediaPlayer" reference installed, but i need to know how it would work...
Attached image(s)
For the following code :
s = 0 ;
for(i=m ; i<=(2*n-1) ; i+=m) {
if(i<=n+1)
{ s+=(i-1)/2 ; }
else
{ s+=(2*n-i+1)/2 ; }
}
I want to change the complexity of the code from O(n) to O(1) . So I wanted to eliminate the for loop . But as the sum "s" stores values like (i-1)/2 or (2*n-i+1)/2 so eliminating the loop involves tedious calculation of floor value of each (i-1)/2 or (2*n-i+1)/2 . It became very difficult for me to do so as I might have derived the wrong formula in sums of floors . Need Changing complexity from O(n) to O(1). Is there any other way to reduce the complexity ? If yes ... then how ?
Assuming that we have :
Code:
int arr2d[rows][columns] ; // Not valid syntax of course ... let be arr2d rows * columns size
for(int i=0; i<rows; i++)
for(int j=0; j<columns; j++)
arr2d[rows][columns] = some_value;
What is the complexity? I believe O(n) and not O(n^2) on this case because if you have 3*3 size you would put 9 elements (from 9 elements input of course)... for each input you have one insertion and that is the meaning. Same as 4*4 size 16 input times 16 insertions .. or 5*5 and so forth...
Giving a dynamic array, we want to pass the array elements from a file. The first number in the file N gives us the array length. They follow N numbers, the actual elements of the array.
Three brothers are going to work in the shop of their father for a time. The shop is doing well and every day generates profit which the three brothers may provide. The brothers agreed that they would divide the total time in three successive parts, not necessarily equal duration, and that each one will be working in the shop during one of these parts and collects the corresponding profit on his behalf. But they want to make a deal fairly, so that one received from three more profit someone else. Specifically, they want to minimize the profit the most favored of the three will receive.
I first created a program that WORKS! with complexity O(n3).
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int sum_array(int* array, int cnt){
int res = 0;
int i;
for ( i = 0; i < cnt ; ++i)
[Code] .....
Let's assume that we have the following input:
10
1 1 8 1 1 3 4 9 5 2
The ouptut should be 15 -> A = 1 + 1 + 8 + 1 + 1 + 3 = 15 , B = 4 + 9 = 13 , C = 5 + 2 = 7.
But the ouptut is 16!
How can I reduce the complexity of my first working! C code?
Suppose you have a switch statement defined like:
switch (parameter) {
case ONE:
case TWO:
// ... other n-2 cases
case N:
doSomething();
break;
default:
somethingElse();
break;
}
What is the complexity of the doSomething()? Is it still equal to the complexity of doSomething() , or the complexity of doSomething() plus the number of case statements? In other words, does the number of case statements a conditional piece of code has count towards the complexity?
This is a round robin execution. with gantt chart. arrival time and burst time. I think there is an error in my formula to get the right answer,i cant resolve it but my program is running. What is the code or the right formula??
#include<stdio.h>
int main(){
int i,j=0,n,time,remain,flag=0,ts;
int sum_wait=0,sum_turnaround=0,at[10],bt[10],rt[10];
int ganttP[50],ganttStartTime[50];
printf("Enter no of Processes : ");
scanf("%d",&n);
remain=n;
[Code] ....
So I'm trying to learn how to use time for time based movement and what not. I did this only knowing the command time(&variable)
time_t timer;
int I, X;
void main() {
time(&timer);
X=timer;
while(I==5) {
[Code] ......
There's probably some other better way to do it... Obviously but for now I see it as putting current time in X.
start while
take in time constantly until I is 5
constantly asking is time>X(preset time 5 seconds ahead)
if it is
display message and add one to I
Why doesn't this display my message after 5 seconds?
how to input with space what i mean is like this, for example:
Product Code: 0000001
Product Name: Mobile Phone
if the users input is like that in the (Product Name)it will skip the next input that is the
Price: Quantity: 100
Manufacturer: Alcatel
something like that...
Heres the code
Code:
if(b<=10)
{/*Start for IF Statement*/
for(a=1;a<=b;a++)
{/*Start of FOR loop*/
printf("
");
printf("Product Code : ");
scanf("%s", &prod_code);
}
[code]....
I used pointer(or is it not?) to make it one part only alphabets and the other one digits. The coding, calculate_charges.c and the open file, customer.txt are attached at the end of the post.
Code:
#include <stdio.h>
#include <string.h>
#define SIZE 3
void trimback(char input[], int strnameindex);
void trimfrnt(char input[], int strnameindex);
}
[code]....
Can we use using declaration within name space scope? is it safe to use it?
I think we can use using declaration for class scope and function scope only.
I want the user to be able to enter a command then a character, like for example: push r....I want the command to be stored in the array command, and the character to be stored in the variable c.
Now I wonder what the best way to get rid of the space is, using scanf or getchar (see below for code, only thing that I changed between the 2 versions is the statement before the comment "get rid of space")? Or maybe it doesnt matter?
Code:
include <stdio.h>
#define MAX 200
void push(char c); // Puts a new element last in queue
char pop(void); // Gets the first element in queue
static char s[MAX];
}
[code]....
