Runtime Complexity in C












0















I have the task to count the steps of count sort and insertion sort. I implemented the counting variable as z and it is stored in the *befehle pointer.



My problem is, that I just can't figure out how to call the functions properly. And the program always tells me, that the result of count sort and insertion sort is different.



The output is a chart with how many steps it took for each n and how long it took.



I don't quite know how much you need to see to help me, so I just post the complete code.



It is mainly the void count_sort function that troubles me.



#include <stdio.h>
#include <stdlib.h>
#include "introprog_complexity_steps_input.h"

const int MAX_VALUE = 5000000;

void count_sort_calculate_counts(int input_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //für x
z=z+1; //für j

for (int x=0;x<=MAX_VALUE;x++){
count_array[x] = 0;
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung der Schleife

}

for (int j = 0; j<len; j++){
z=z+1; // Vergleich
z=z+1; // Inkrementierung d. Schleife
count_array[input_array[j]]++;
z=z+1; //Inkrementierung
}
*befehle = z;
}

void count_sort_write_output_array(int output_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //Allozierung v. Speicher in Count_Sort
z=z+1; //Für k
z=z+1; //Zuweisung v. j
z=z+1; //Zuweisung v. i

int k = 0;
for (int j = 0;j <=len;j++){
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife
for(int i = 0; i<count_array[j]; i++){
z=z+1; // Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife

output_array[k] = j;

z=z+1; // Zuweisung
z=z+1; // Inkrementierung v. k

k=k+1;
}
}
*befehle=z;
}


void count_sort(int array, int len, int* befehle) {

int* count_array = malloc(sizeof(int) * MAX_VALUE);

int input_array[len];
int output_array[len];



count_sort_calculate_counts(input_array, len, count_array, befehle);
count_sort_write_output_array(output_array, len, count_array, befehle);
free(count_array);
}


void insertion_sort(int array, int len, int* befehle) {
//muss implementiert werden
int z=0;
int j, i, key;
z=z+1; // j Zuweisung
for (j=1; j<len; j++) {
z=z+1; // Vergleich
z=z+1; // Inkrementierung v. j
key = array[j];
z=z+1; // Zuweisung
i=j-1;
z=z+1; //Inkrementierung

while (i>=0 && array[i] > key) {
z=z+1; //Vergleich 1
//z=z+1; //Vergleich 2
array[i+1] = array[i];
z=z+1; //Zuweisung
i = i-1;
z=z+1; //Inkrementierung
}
array[i+1] = key;
z=z+1; //Zuweisung
}
*befehle=z;
}


int main(int argc, char *argv) {

const int WERTE = {1000,2000,3000,4000,5000};
const int LEN_WERTE = 5;
const int LEN_ALGORITHMEN = 2;

int rc = 0;
long befehle_array[LEN_ALGORITHMEN][LEN_WERTE];
double laufzeit_array[LEN_ALGORITHMEN][LEN_WERTE];

for(int j = 0; j < LEN_WERTE; ++j)
{
int n = WERTE[j];

//reserviere Speicher für Arrays der Länge n
int* array_countsort = malloc(sizeof(int) * n);
int* array_insertionsort = malloc(sizeof(int) * n);

//fülle array_countsort mit Zufallswerten ..
fill_array_randomly(array_countsort, n, MAX_VALUE);
//.. und kopiere die erzeugten Werte in das Array array_insertionsort
copy_array_elements(array_insertionsort, array_countsort, n);

//teste ob beide Arrays auch wirklich die gleichen Werte enthalten
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Eingaben für beide Algorithmen müssen für die Vergleichbarkeit gleich sein!n");
return -1;
}

for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
int anzahl_befehle = 0;

start_timer();

//Aufruf der entsprechenden Sortieralgorithmen
if(i==0)
{
count_sort(array_countsort, n, &anzahl_befehle);
}
else if(i==1)
{
insertion_sort(array_insertionsort, n, &anzahl_befehle);
}

//speichere die Laufzeit sowie die Anzahl benötigter Befehle
laufzeit_array[i][j] = end_timer();
befehle_array[i][j] = anzahl_befehle;
}

//teste ob die Ausgabe beider Algorithmen gleich ist
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Arrays sind nicht gleich. Eines muss (falsch) sortiert worden sein!n");
rc = -1;
}

//gib den Speicherplatz wieder frei
free(array_countsort);
free(array_insertionsort);
}

//Ausgabe der Anzahl ausgeführter Befehle sowie der gemessenen Laufzeiten (in Millisekunden)
printf("Parameter MAX_VALUE hat den Wert %dn", MAX_VALUE);
printf("t %32s %32s n", "Countsort","Insertionsort");
printf("%8s t %16s %16s t %16s %16s n", "n","Befehle", "Laufzeit","Befehle","Laufzeit");

for(int j = 0; j < LEN_WERTE; ++j)
{
printf("%8d t ",WERTE[j]);
for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
printf("%16ld %16.4f t ", befehle_array[i][j], laufzeit_array[i][j]);
}
printf("n");
}

return rc;
}









share|improve this question

























  • There are some advantages to z++ and z += 2;. Why do you expect two different sorts to produce the same counting result? If you used quick sort or heap sort or merge sort, the counts would be different each time. It isn't clear to me what you're counting. Should you be using *befehle += z; at the ends of functions, rather than just overwriting the count?

    – Jonathan Leffler
    Nov 28 '18 at 18:03
















0















I have the task to count the steps of count sort and insertion sort. I implemented the counting variable as z and it is stored in the *befehle pointer.



My problem is, that I just can't figure out how to call the functions properly. And the program always tells me, that the result of count sort and insertion sort is different.



The output is a chart with how many steps it took for each n and how long it took.



I don't quite know how much you need to see to help me, so I just post the complete code.



It is mainly the void count_sort function that troubles me.



#include <stdio.h>
#include <stdlib.h>
#include "introprog_complexity_steps_input.h"

const int MAX_VALUE = 5000000;

void count_sort_calculate_counts(int input_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //für x
z=z+1; //für j

for (int x=0;x<=MAX_VALUE;x++){
count_array[x] = 0;
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung der Schleife

}

for (int j = 0; j<len; j++){
z=z+1; // Vergleich
z=z+1; // Inkrementierung d. Schleife
count_array[input_array[j]]++;
z=z+1; //Inkrementierung
}
*befehle = z;
}

void count_sort_write_output_array(int output_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //Allozierung v. Speicher in Count_Sort
z=z+1; //Für k
z=z+1; //Zuweisung v. j
z=z+1; //Zuweisung v. i

int k = 0;
for (int j = 0;j <=len;j++){
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife
for(int i = 0; i<count_array[j]; i++){
z=z+1; // Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife

output_array[k] = j;

z=z+1; // Zuweisung
z=z+1; // Inkrementierung v. k

k=k+1;
}
}
*befehle=z;
}


void count_sort(int array, int len, int* befehle) {

int* count_array = malloc(sizeof(int) * MAX_VALUE);

int input_array[len];
int output_array[len];



count_sort_calculate_counts(input_array, len, count_array, befehle);
count_sort_write_output_array(output_array, len, count_array, befehle);
free(count_array);
}


void insertion_sort(int array, int len, int* befehle) {
//muss implementiert werden
int z=0;
int j, i, key;
z=z+1; // j Zuweisung
for (j=1; j<len; j++) {
z=z+1; // Vergleich
z=z+1; // Inkrementierung v. j
key = array[j];
z=z+1; // Zuweisung
i=j-1;
z=z+1; //Inkrementierung

while (i>=0 && array[i] > key) {
z=z+1; //Vergleich 1
//z=z+1; //Vergleich 2
array[i+1] = array[i];
z=z+1; //Zuweisung
i = i-1;
z=z+1; //Inkrementierung
}
array[i+1] = key;
z=z+1; //Zuweisung
}
*befehle=z;
}


int main(int argc, char *argv) {

const int WERTE = {1000,2000,3000,4000,5000};
const int LEN_WERTE = 5;
const int LEN_ALGORITHMEN = 2;

int rc = 0;
long befehle_array[LEN_ALGORITHMEN][LEN_WERTE];
double laufzeit_array[LEN_ALGORITHMEN][LEN_WERTE];

for(int j = 0; j < LEN_WERTE; ++j)
{
int n = WERTE[j];

//reserviere Speicher für Arrays der Länge n
int* array_countsort = malloc(sizeof(int) * n);
int* array_insertionsort = malloc(sizeof(int) * n);

//fülle array_countsort mit Zufallswerten ..
fill_array_randomly(array_countsort, n, MAX_VALUE);
//.. und kopiere die erzeugten Werte in das Array array_insertionsort
copy_array_elements(array_insertionsort, array_countsort, n);

//teste ob beide Arrays auch wirklich die gleichen Werte enthalten
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Eingaben für beide Algorithmen müssen für die Vergleichbarkeit gleich sein!n");
return -1;
}

for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
int anzahl_befehle = 0;

start_timer();

//Aufruf der entsprechenden Sortieralgorithmen
if(i==0)
{
count_sort(array_countsort, n, &anzahl_befehle);
}
else if(i==1)
{
insertion_sort(array_insertionsort, n, &anzahl_befehle);
}

//speichere die Laufzeit sowie die Anzahl benötigter Befehle
laufzeit_array[i][j] = end_timer();
befehle_array[i][j] = anzahl_befehle;
}

//teste ob die Ausgabe beider Algorithmen gleich ist
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Arrays sind nicht gleich. Eines muss (falsch) sortiert worden sein!n");
rc = -1;
}

//gib den Speicherplatz wieder frei
free(array_countsort);
free(array_insertionsort);
}

//Ausgabe der Anzahl ausgeführter Befehle sowie der gemessenen Laufzeiten (in Millisekunden)
printf("Parameter MAX_VALUE hat den Wert %dn", MAX_VALUE);
printf("t %32s %32s n", "Countsort","Insertionsort");
printf("%8s t %16s %16s t %16s %16s n", "n","Befehle", "Laufzeit","Befehle","Laufzeit");

for(int j = 0; j < LEN_WERTE; ++j)
{
printf("%8d t ",WERTE[j]);
for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
printf("%16ld %16.4f t ", befehle_array[i][j], laufzeit_array[i][j]);
}
printf("n");
}

return rc;
}









share|improve this question

























  • There are some advantages to z++ and z += 2;. Why do you expect two different sorts to produce the same counting result? If you used quick sort or heap sort or merge sort, the counts would be different each time. It isn't clear to me what you're counting. Should you be using *befehle += z; at the ends of functions, rather than just overwriting the count?

    – Jonathan Leffler
    Nov 28 '18 at 18:03














0












0








0








I have the task to count the steps of count sort and insertion sort. I implemented the counting variable as z and it is stored in the *befehle pointer.



My problem is, that I just can't figure out how to call the functions properly. And the program always tells me, that the result of count sort and insertion sort is different.



The output is a chart with how many steps it took for each n and how long it took.



I don't quite know how much you need to see to help me, so I just post the complete code.



It is mainly the void count_sort function that troubles me.



#include <stdio.h>
#include <stdlib.h>
#include "introprog_complexity_steps_input.h"

const int MAX_VALUE = 5000000;

void count_sort_calculate_counts(int input_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //für x
z=z+1; //für j

for (int x=0;x<=MAX_VALUE;x++){
count_array[x] = 0;
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung der Schleife

}

for (int j = 0; j<len; j++){
z=z+1; // Vergleich
z=z+1; // Inkrementierung d. Schleife
count_array[input_array[j]]++;
z=z+1; //Inkrementierung
}
*befehle = z;
}

void count_sort_write_output_array(int output_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //Allozierung v. Speicher in Count_Sort
z=z+1; //Für k
z=z+1; //Zuweisung v. j
z=z+1; //Zuweisung v. i

int k = 0;
for (int j = 0;j <=len;j++){
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife
for(int i = 0; i<count_array[j]; i++){
z=z+1; // Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife

output_array[k] = j;

z=z+1; // Zuweisung
z=z+1; // Inkrementierung v. k

k=k+1;
}
}
*befehle=z;
}


void count_sort(int array, int len, int* befehle) {

int* count_array = malloc(sizeof(int) * MAX_VALUE);

int input_array[len];
int output_array[len];



count_sort_calculate_counts(input_array, len, count_array, befehle);
count_sort_write_output_array(output_array, len, count_array, befehle);
free(count_array);
}


void insertion_sort(int array, int len, int* befehle) {
//muss implementiert werden
int z=0;
int j, i, key;
z=z+1; // j Zuweisung
for (j=1; j<len; j++) {
z=z+1; // Vergleich
z=z+1; // Inkrementierung v. j
key = array[j];
z=z+1; // Zuweisung
i=j-1;
z=z+1; //Inkrementierung

while (i>=0 && array[i] > key) {
z=z+1; //Vergleich 1
//z=z+1; //Vergleich 2
array[i+1] = array[i];
z=z+1; //Zuweisung
i = i-1;
z=z+1; //Inkrementierung
}
array[i+1] = key;
z=z+1; //Zuweisung
}
*befehle=z;
}


int main(int argc, char *argv) {

const int WERTE = {1000,2000,3000,4000,5000};
const int LEN_WERTE = 5;
const int LEN_ALGORITHMEN = 2;

int rc = 0;
long befehle_array[LEN_ALGORITHMEN][LEN_WERTE];
double laufzeit_array[LEN_ALGORITHMEN][LEN_WERTE];

for(int j = 0; j < LEN_WERTE; ++j)
{
int n = WERTE[j];

//reserviere Speicher für Arrays der Länge n
int* array_countsort = malloc(sizeof(int) * n);
int* array_insertionsort = malloc(sizeof(int) * n);

//fülle array_countsort mit Zufallswerten ..
fill_array_randomly(array_countsort, n, MAX_VALUE);
//.. und kopiere die erzeugten Werte in das Array array_insertionsort
copy_array_elements(array_insertionsort, array_countsort, n);

//teste ob beide Arrays auch wirklich die gleichen Werte enthalten
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Eingaben für beide Algorithmen müssen für die Vergleichbarkeit gleich sein!n");
return -1;
}

for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
int anzahl_befehle = 0;

start_timer();

//Aufruf der entsprechenden Sortieralgorithmen
if(i==0)
{
count_sort(array_countsort, n, &anzahl_befehle);
}
else if(i==1)
{
insertion_sort(array_insertionsort, n, &anzahl_befehle);
}

//speichere die Laufzeit sowie die Anzahl benötigter Befehle
laufzeit_array[i][j] = end_timer();
befehle_array[i][j] = anzahl_befehle;
}

//teste ob die Ausgabe beider Algorithmen gleich ist
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Arrays sind nicht gleich. Eines muss (falsch) sortiert worden sein!n");
rc = -1;
}

//gib den Speicherplatz wieder frei
free(array_countsort);
free(array_insertionsort);
}

//Ausgabe der Anzahl ausgeführter Befehle sowie der gemessenen Laufzeiten (in Millisekunden)
printf("Parameter MAX_VALUE hat den Wert %dn", MAX_VALUE);
printf("t %32s %32s n", "Countsort","Insertionsort");
printf("%8s t %16s %16s t %16s %16s n", "n","Befehle", "Laufzeit","Befehle","Laufzeit");

for(int j = 0; j < LEN_WERTE; ++j)
{
printf("%8d t ",WERTE[j]);
for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
printf("%16ld %16.4f t ", befehle_array[i][j], laufzeit_array[i][j]);
}
printf("n");
}

return rc;
}









share|improve this question
















I have the task to count the steps of count sort and insertion sort. I implemented the counting variable as z and it is stored in the *befehle pointer.



My problem is, that I just can't figure out how to call the functions properly. And the program always tells me, that the result of count sort and insertion sort is different.



The output is a chart with how many steps it took for each n and how long it took.



I don't quite know how much you need to see to help me, so I just post the complete code.



It is mainly the void count_sort function that troubles me.



#include <stdio.h>
#include <stdlib.h>
#include "introprog_complexity_steps_input.h"

const int MAX_VALUE = 5000000;

void count_sort_calculate_counts(int input_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //für x
z=z+1; //für j

for (int x=0;x<=MAX_VALUE;x++){
count_array[x] = 0;
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung der Schleife

}

for (int j = 0; j<len; j++){
z=z+1; // Vergleich
z=z+1; // Inkrementierung d. Schleife
count_array[input_array[j]]++;
z=z+1; //Inkrementierung
}
*befehle = z;
}

void count_sort_write_output_array(int output_array, int len, int count_array, int* befehle) {
//muss implementiert werden
int z = 0;
z=z+1; //Allozierung v. Speicher in Count_Sort
z=z+1; //Für k
z=z+1; //Zuweisung v. j
z=z+1; //Zuweisung v. i

int k = 0;
for (int j = 0;j <=len;j++){
z=z+1; //Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife
for(int i = 0; i<count_array[j]; i++){
z=z+1; // Vergleich d. Schleife
z=z+1; //Inkrementierung d. Schleife

output_array[k] = j;

z=z+1; // Zuweisung
z=z+1; // Inkrementierung v. k

k=k+1;
}
}
*befehle=z;
}


void count_sort(int array, int len, int* befehle) {

int* count_array = malloc(sizeof(int) * MAX_VALUE);

int input_array[len];
int output_array[len];



count_sort_calculate_counts(input_array, len, count_array, befehle);
count_sort_write_output_array(output_array, len, count_array, befehle);
free(count_array);
}


void insertion_sort(int array, int len, int* befehle) {
//muss implementiert werden
int z=0;
int j, i, key;
z=z+1; // j Zuweisung
for (j=1; j<len; j++) {
z=z+1; // Vergleich
z=z+1; // Inkrementierung v. j
key = array[j];
z=z+1; // Zuweisung
i=j-1;
z=z+1; //Inkrementierung

while (i>=0 && array[i] > key) {
z=z+1; //Vergleich 1
//z=z+1; //Vergleich 2
array[i+1] = array[i];
z=z+1; //Zuweisung
i = i-1;
z=z+1; //Inkrementierung
}
array[i+1] = key;
z=z+1; //Zuweisung
}
*befehle=z;
}


int main(int argc, char *argv) {

const int WERTE = {1000,2000,3000,4000,5000};
const int LEN_WERTE = 5;
const int LEN_ALGORITHMEN = 2;

int rc = 0;
long befehle_array[LEN_ALGORITHMEN][LEN_WERTE];
double laufzeit_array[LEN_ALGORITHMEN][LEN_WERTE];

for(int j = 0; j < LEN_WERTE; ++j)
{
int n = WERTE[j];

//reserviere Speicher für Arrays der Länge n
int* array_countsort = malloc(sizeof(int) * n);
int* array_insertionsort = malloc(sizeof(int) * n);

//fülle array_countsort mit Zufallswerten ..
fill_array_randomly(array_countsort, n, MAX_VALUE);
//.. und kopiere die erzeugten Werte in das Array array_insertionsort
copy_array_elements(array_insertionsort, array_countsort, n);

//teste ob beide Arrays auch wirklich die gleichen Werte enthalten
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Eingaben für beide Algorithmen müssen für die Vergleichbarkeit gleich sein!n");
return -1;
}

for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
int anzahl_befehle = 0;

start_timer();

//Aufruf der entsprechenden Sortieralgorithmen
if(i==0)
{
count_sort(array_countsort, n, &anzahl_befehle);
}
else if(i==1)
{
insertion_sort(array_insertionsort, n, &anzahl_befehle);
}

//speichere die Laufzeit sowie die Anzahl benötigter Befehle
laufzeit_array[i][j] = end_timer();
befehle_array[i][j] = anzahl_befehle;
}

//teste ob die Ausgabe beider Algorithmen gleich ist
if(!check_equality_of_arrays(array_countsort, array_insertionsort, n))
{
printf("Die Arrays sind nicht gleich. Eines muss (falsch) sortiert worden sein!n");
rc = -1;
}

//gib den Speicherplatz wieder frei
free(array_countsort);
free(array_insertionsort);
}

//Ausgabe der Anzahl ausgeführter Befehle sowie der gemessenen Laufzeiten (in Millisekunden)
printf("Parameter MAX_VALUE hat den Wert %dn", MAX_VALUE);
printf("t %32s %32s n", "Countsort","Insertionsort");
printf("%8s t %16s %16s t %16s %16s n", "n","Befehle", "Laufzeit","Befehle","Laufzeit");

for(int j = 0; j < LEN_WERTE; ++j)
{
printf("%8d t ",WERTE[j]);
for(int i = 0; i < LEN_ALGORITHMEN; ++i)
{
printf("%16ld %16.4f t ", befehle_array[i][j], laufzeit_array[i][j]);
}
printf("n");
}

return rc;
}






c runtime






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 28 '18 at 16:32









Jonathan Leffler

572k926861037




572k926861037










asked Nov 28 '18 at 11:25









LennoxLennox

183




183













  • There are some advantages to z++ and z += 2;. Why do you expect two different sorts to produce the same counting result? If you used quick sort or heap sort or merge sort, the counts would be different each time. It isn't clear to me what you're counting. Should you be using *befehle += z; at the ends of functions, rather than just overwriting the count?

    – Jonathan Leffler
    Nov 28 '18 at 18:03



















  • There are some advantages to z++ and z += 2;. Why do you expect two different sorts to produce the same counting result? If you used quick sort or heap sort or merge sort, the counts would be different each time. It isn't clear to me what you're counting. Should you be using *befehle += z; at the ends of functions, rather than just overwriting the count?

    – Jonathan Leffler
    Nov 28 '18 at 18:03

















There are some advantages to z++ and z += 2;. Why do you expect two different sorts to produce the same counting result? If you used quick sort or heap sort or merge sort, the counts would be different each time. It isn't clear to me what you're counting. Should you be using *befehle += z; at the ends of functions, rather than just overwriting the count?

– Jonathan Leffler
Nov 28 '18 at 18:03





There are some advantages to z++ and z += 2;. Why do you expect two different sorts to produce the same counting result? If you used quick sort or heap sort or merge sort, the counts would be different each time. It isn't clear to me what you're counting. Should you be using *befehle += z; at the ends of functions, rather than just overwriting the count?

– Jonathan Leffler
Nov 28 '18 at 18:03












0






active

oldest

votes











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53518356%2fruntime-complexity-in-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53518356%2fruntime-complexity-in-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Contact image not getting when fetch all contact list from iPhone by CNContact

count number of partitions of a set with n elements into k subsets

A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks