How can I display the input user of this code. I am using bubble sort but I cannot display the input.
Here's my code:
#include <stdio.h>
int main(void) {
int array[100];
int count;
for(int i = 0; i < 100; i++) {
printf("Enter number %d: ", i + 1);
scanf("%d", &array[i]);
if(array[i] == 0) break;
}
int size = sizeof(array) / sizeof(array[0]);
for(int i = 0; i < size - 1; ++i) {
int swapped = 0;
for(int ctr = 0; ctr < size - i - 1; ++ctr) {
if(array[ctr] > array[ctr + 1]) {
int temp = array[ctr];
array[ctr] = array[ctr + 1];
array[ctr + 1] = temp;
swapped = 1;
}
}
if(swapped == 0) {
break;
}
}
printf("[");
for(int i = 0; i < size; i++) {
if(i == size - 1) {
printf("%d", array[i]);
}
else {
printf("%d, ", array[i]);
}
}
printf("]");
return 0;
}
Here's the output:
Enter·number·1:·1
Enter·number·2:·2
Enter·number·3:·3
Enter·number·4:·4
Enter·number·5:·5
Enter·number·6:·0
[1,·2,·3,·4,·5]
I cannot display the 1 2 3 4 5.... How can I display?
The calculation of the size doesn't work like you did it. sizeof(array) will give you the total amount of of bytes of the array (of all 100 spaces). So your size will allway be 100.
Instead of calculating the size it's easyer to count the inputs like this:
int size=0;
for(int i = 0; i < 100; i++) {
printf("Enter number %d: ", i + 1);
scanf("%d", &array[i]);
if(array[i] == 0) break;
size++;
}
The user can enter i values in the array due to the break statement within the for loop
for(int i = 0; i < 100; i++) {
printf("Enter number %d: ", i + 1);
scanf("%d", &array[i]);
if(array[i] == 0) break;
}
So you need to keep this value and use it the following for loops because it can differ from the value of the expression sizeof(array) / sizeof(array[0])
Related
I want to store elements of maximum and minimum frequency in the arr2 array if there are more than one element of same frequency then both the elements should be stored ? But it is showing wrong results and i am not able to find what is the err. Can anyone help me with this. Any kind of help would be greatly appreciated.
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
int arr[n];
for (int i = 0; i < n; i++)
{
scanf("%d", &arr[i]);
}
int arr2[n];
int prevcount = 0;
int k = 0;
// for finding max element
for (int i = 0; i < n; i++)
{
int count = 0;
//counting the number of times it has occured
for (int j = 0; j < n; j++)
{
if (arr[i] == arr[j])
{
count++;
}
}
// checking if the same element was not there in the new array
for (int i = 0; i < k; i++)
{
if (arr[i] == arr[k])
{
goto nextit;
}
}
//it will update the kth element if the count is greater than the prev count
if (prevcount < count)
{
arr2[k] = arr[i];
}
//if these both are same but the number is different then will iterate k by 1 and store that element as well
else if (prevcount == count)
{
k++;
arr2[k] = arr[i];
}
prevcount = count;
nextit:
}
// for finding min element
prevcount = 1000;
for (int i = 0; i < n; i++)
{
int count = 0;
for (int j = 0; j < n; j++)
{
if (arr[i] == arr[j])
{
count++;
}
}
// checking if the same element was not there in the new array if there is then go to the next iteration
for (int i = 0; i < k; i++)
{
if (arr[i] == arr[k])
{
goto nextit2;
}
}
if (prevcount > count)
{
arr2[k] = arr[i];
}
else if (prevcount == count)
{
k++;
arr2[k] = arr[i];
}
prevcount = count;
nextit2:
}
for (int i = 0; i < k; i++)
{
printf("%d ", arr2[i]);
}
return 0;
}
As #SparKot suggests, sorting the array makes the problem simple. Would you please try:
#include <stdio.h>
#include <stdlib.h>
// compare values numerically
int numeric(const void *a, const void *b)
{
return (*(int *)a < *(int *)b) ? -1 : (*(int *)a > *(int *)b);
}
int main()
{
int n, i, j;
int *arr; // input array
int *count; // count frequency: initialized to 0's by calloc
int min = 0; // minimum occurrences
int max = 0; // maximum occurrences
scanf("%d", &n);
if (NULL == (arr = malloc(n * sizeof(int)))) {
perror("malloc");
exit(1);
}
if (NULL == (count = calloc(n, sizeof(int)))) {
perror("calloc");
exit(1);
}
for (i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
qsort(arr, n, sizeof(int), numeric);
// count the length of sequence of the same numbers
for (i = 0; i < n; i++) {
for (j = 0; i + j < n && arr[i] == arr[i + j]; j++) {
;
}
count[i] = j; // i'th element has length j
i += j - 1; // jump to next number
}
// find minimum and maximum frequencies
for (i = 0; i < n; i++) {
if (count[i]) {
if (min == 0 || count[i] < min) min = count[i];
if (max == 0 || count[i] > max) max = count[i];
}
}
// report the result
for (i = 0; i < n; i++) {
if (count[i] == min) {
printf("min frequency %d value %d\n", count[i], arr[i]);
}
if (count[i] == max) {
printf("max frequency %d value %d\n", count[i], arr[i]);
}
}
return 0;
}
Sample input (n=10):
6
1
2
5
1
2
3
1
3
6
Output:
max frequency 3 value 1
min frequency 1 value 5
I am having trouble achieving the wanted results. The program should ask for 20 inputs and then go over each to see if they appear more than once. Then only print out those that appeared once.
However currently my program prints out random numbers that are not inputted.
For example:
array = {10,10,11,12,10,10,10.....,10} should return 11 and 12
#include <stdio.h>
void main() {
int count, size=20, array[size], newArr[size];
int number=0;
for(count = 0; count < size; count++) {
// Ask user for input until 20 correct inputs.
printf("\nAnna %d. luku > ", count+1);
scanf("%d", &number);
if( (number > 100) || (number < 10) ) {
while(1) {
number = 0;
printf("Ei kelpaa.\n");//"Is not valid"
printf("Yrita uudelleen > ");//"Try again >"
scanf("%d", &number);
if ( (number <= 100) && (number >= 10) ) {
break;
}
}
}
array[count] = number;
}
for(int i=0; i < size; i++) {
for(int j=0; j<size; j++){
if(array[i] == array[j]){
size--;
break;
} else {
// if not duplicate add to the new array
newArr[i] == array[j];
}
}
}
// print out all the elements of the new array
for(int k=0; k<size; k++) {
printf("%d\n", newArr[k]);
}
}
You don't need the newArr here, or the separate output loop. Only keep a count that you reset to zero at the beginning of the outer loop, and increase in the inner loop if you find a duplicate.
Once the inner loop is finished, and the counter is 1 then you don't have any duplicates and you print the value.
In code perhaps something like:
for (unsigned i = 0; i < size; ++i)
{
unsigned counter = 0;
for (unsigned j = 0; j < size; ++j)
{
if (array[i] == array[j])
{
++counter;
}
}
if (counter == 1)
{
printf("%d\n", array[i]);
}
}
Note that the above is a pretty naive and brute-force way to deal with it, and that it will not perform very well for larger array sizes.
Then one could implement a hash-table, where the value is the key, and the count is the data.
Each time you read a value you increase the data for that value.
Once done iterate over the map and print all values whose data (counter) is 1.
Use functions!!
Use proper types for indexes (size_t).
void printdistinct(const int *arr, size_t size)
{
int dist;
for(size_t s = 0; s < size; s++)
{
int val = arr[s];
dist = 1;
for(size_t n = 0; n < size; n++)
{
if(s != n)
if(val == arr[n]) {dist = 0; break;}
}
if(dist) printf("%d ", val);
}
printf("\n");
}
int main(void)
{
int test[] = {10,10,11,12,10,10,10,10};
printdistinct(test, sizeof(test)/sizeof(test[0]));
fflush(stdout);
}
https://godbolt.org/z/5bKfdn9Wv
This is how I did it and it should work for your:
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <stdarg.h>
void printdistinct(const int *muis, size_t size);
int main()
{
int loop=20,i,muis[20],monesko=0;
for(i=0; i<loop; i++){
monesko++;
printf ("Anna %d. luku: \n",monesko);
scanf("%d", &muis[i]);
if (muis[i]<10 || muis[i]>100){
printf("Ei kelpaa!\n");
muis[i] = muis[i + 1];
printf("YRITÄ UUDELLEEN:\n ");
scanf("%d", &muis[i]);
}
}
printdistinct(muis, sizeof(muis)/sizeof(muis[0]));
fflush(stdout);
return 0;
}
void printdistinct(const int *muis, size_t size)
{
for(size_t s = 0; s < size; s++)
{
int a = muis[s];
int testi = 1;
for(size_t n = 0; n < size; n++){
if(s != n) {
if(a == muis[n]){
testi = 0;
break;
}
}
}
if(testi) {
printf("%d \n", a);
}
testi = 1;
}
printf("\n");
}
This approach uses some memory to keep track of which elements are duplicates. The memory cost is higher, but the processor time cost is lower. These differences will become significant at higher values of size.
char* duplicate = calloc(size, 1); // values are initialized to zero
for (unsigned i = 0; i < size; ++i)
{
if(!duplicate[i]) // skip any value that's known to be a duplicate
{
for (unsigned j = i + 1; j < size; ++j) // only look at following values
{
if (array[i] == array[j])
{
duplicate[i] = 1;
duplicate[j] = 1; // all duplicates will be marked
}
}
if (!duplicate[i])
{
printf("%d\n", array[i]);
}
}
}
What you can do is you can initialize a hashmap that will help you store the unique elements. Once you start iterating the array you check for that element in the hashmap. If it is not present in the hashmap add it to the hashmap. If it is already present keep iterating.
This way you would not have to iterate the loop twice. Your time complexity of the algorithm will be O(n).
unordered_map < int, int > map;
for (int i = 0; i < size; i++) {
// Check if present in the hashmap
if (map.find(arr[i]) == map.end()) {
// Insert the element in the hash map
map[arr[i]] = arr[i];
cout << arr[i] << " ";
}
}
So I have this code in C:
#include <stdio.h>
int main() {
int a[9], number_of_elements;
printf("Enter 10 numbers: \n");
for (int i = 0; ((i < 10) && (a[i] != 0)); i++) {
scanf_s("%d", &a[i]);
if (a[i] == 0) {
number_of_elements = i;
break;
}
if (a[i] != 0) {
number_of_elements = i + 1;
}
}
printf("There is %d elements.\n", number_of_elements);
return 0;
}
As you can see, I have to enter max 10 elements into an array, or stop entering when you enter 0, and after that I have to print all elements of that array and how many elements array have. I sorted all things except printing all elements of the array, can anyone help me how to do that?
You should use another for loop, like:
for(i=0;i<number_of_elements;i++)
{
printf("%d ",a[i]);
}
After implementing #rsp's code, I fixed my code.
#include <stdio.h>
int main() {
int a[10], number_of_elements;
printf("Enter 10 numbers: \n");
for (int i = 0; ((i < 10) && (a[i] != 0)); i++) {
scanf_s("%d", &a[i]);
if (a[i] == 0) {
number_of_elements = i;
break;
}
if (a[i] != 0) {
number_of_elements = i + 1;
}
}
for (int i = 0; i<number_of_elements; i++)
{
printf("%d ", a[i]);
}
printf("\nThere are %d elements.\n", number_of_elements);
return 0;
}
Tnx everyone for help and tnx #George for being so funny :)
I'm trying to allocate some memory using malloc in my program (I don't have much experience with malloc as I am just starting to learn how to use it). The program I've created works but I don't think I've used malloc correctly and I want to learn how to use it correctly.
The basic idea of my program is it takes 3 lines of input. The first line being whether you want to sort it by odd or even, the second line being how large your array is and the third line being the integers in the array.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* sort;
int n;
int* ar;
int i;
void test()
{
int temp;
int j = 1;
if (strcmp(sort, "odd") == 0) {
for (i = 0; i < n;) {
if (j != n) {
if (ar[i] % 2 != 0) {
if (ar[j] % 2 != 0) {
if (ar[j] < ar[i]) {
temp = ar[i];
ar[i] = ar[j];
ar[j] = temp;
j++;
}
else {
j++;
}
}
else {
j++;
}
}
else {
j++;
i++;
}
}
else {
i++;
j = i + 1;
}
}
}
if (strcmp(sort, "even") == 0) {
for (i = 0; i < n; i++) {
if (j != n) {
if (ar[i] % 2 == 0) {
if (ar[j] % 2 == 0) {
if (ar[j] < ar[i]) {
temp = ar[i];
ar[i] = ar[j];
ar[j] = temp;
j++;
}
else {
j++;
}
}
else {
j++;
}
}
else {
j++;
i++;
}
}
else {
i++;
j = i + 1;
}
}
}
}
void main()
{
ar = malloc(sizeof(int) * 10);
sort = malloc(sizeof(char) + 5);
printf("Enter odd or even\n");
scanf("%s", sort);
printf("Enter the size of the array \n");
scanf("%d", &n);
printf("Enter the elements of the array \n");
for (i = 0; i < n; i++) {
scanf("%d", &ar[i]);
}
test();
for (i = 0; i < n; i++) {
printf("%d ", ar[i]);
}
}
You need to wait until they tell you the size of the array, and then use malloc() to allocate an array that size.
You don't need to use malloc for sort, you can just declare that as an ordinary array. In general, you use malloc() either when you need to allocate a dynamic number of items, or when the size of the item is dynamic, you don't need it for a single item with a fixed size.
int *ar;
char sort[5];
void main()
{
printf("Enter odd or even\n");
scanf("%s", sort);
printf("Enter the size of the array \n");
scanf("%d", &n);
ar = malloc(n * sizeof(int));
printf("Enter the elements of the array \n");
for (i = 0; i < n; i++) {
scanf("%d", &ar[i]);
}
test();
for (i = 0; i < n; i++) {
printf("%d ", ar[i]);
}
}
I am writing a program that arranges an array in nondecreasing order; then, it inserts a value into the sequence. I can easily get numbers in the beginning and the middle of the array, but whenever I add a number that should go at the end, I keep getting 0. Where am I going wrong?
#include <stdio.h>
int main()
{
int array[10];
int i, j, n, m, temp, key, pos;
printf("Enter number of elements:\n");
scanf("%d", &n);
printf("Enter the elements:\n");
for (i = 0; i < n; i++)
{
scanf("%d", &array[i]);
}
printf("Input array elements:\n");
for (i = 0; i < n; i++)
{
printf("%d\n", array[i]);
}
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
if (array[i] > array[j])
{
temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
}
printf("Sorted list is\n");
for (i = 0; i < n; i++)
{
printf("%d\n", array[i]);
}
printf("Enter the element to be inserted X:\n");
scanf("%d", &key);
for (i = 0; ; i++)
{
if (key < array[i])
{
pos = i;
break;
}
}
m = n - pos + 1 ;
for (i = 0; i <= m; i++)
{
array[n - i + 2] = array[n - i + 1] ;
}
array[pos] = key;
printf("Final list is:\n");
for (i = 0; i < n + 1; i++)
{
printf("%d\n", array[i]);
}
}
If the number to be entered is larger than all elements the following loop poses an issue.
...
for (i = 0; ; i++)
{
if (key < array[i])
{
pos = i;
break;
}
}
If the number is largest,then the pos will be junk and not n.The default value of an array is junk.
Replace it with this.
for (i = 0;i<n ; i++)
{
if (key < array[i])
{
break;
}
}
pos = i;
...