Replacing repeated values in an array in C - c

I'm coding a function in C to check if an array has any duplicate value, and if so replace it by any of the non-present ones. The array consists of numbers shuffled from 1 to Nmax:
unsigned char * NotRepeated (unsigned char *arr){
unsigned char i, j, count, k, notrepeated[Nmax];
k = 0;
for (i = 0; i < Nmax ; i++){
count = 0;
for (j = 0; j < Nmax; j++){
if (arr[j] != i + 1){
count++;
}
if (count == Nmax){
notrepeated[k] = i + 1;
k++;
}
}
}
k = 0;
for (i = 0; i < Nmax - 1; i++){
for (j = i + 1; j < Nmax; j++){
if (arr[i] == arr[j]){
arr[j] = notrepeated[k];
k++;
}
}
}
return arr
}
If I print the array notrepeated[k] I get almost all the array filled when the original array arr[j] seldomly has more than 2 repeated figures.
What am I doing wrong?
Thanks!

Related

Sorting a matrix

Im trying to sort a matrix by the sum of its row's digits, from highest to lowest. I dont know if i explained that correctly so here's some photos explaining it.
This is what my code outputs. Basically, it asks you for m and n, which are the dimensions of the matrix. In this example it's a 3x4, 3 rows and 4 columns. Then, the matrix should be sorted by rows, by the sum of row's digits. Which means, instead of what's being outputted in the picture above, the correct result should be this:
I have no idea how to sort this from highest to lowest, i have been trying for hours to no avail.
Here's my code:
#include <stdio.h>
#define N 30
void main(){
double a[N][N], s[N], p;
int i, j, m, n, max;
while(1){
printf("\nm, n? ");
scanf("%d%d", &m, &n);
if(m <= 0 || m > N || n <=0 || n > N)
break;
for(i = 0; i < m; i++){
printf("%2d. row? ", i+1);
for(j = 0; j < n; scanf("%lf", &a[i][j++]));
}
for(i = 0; i < m; i++)
for(s[i] = j = 0; j < n; s[i] += a[i][j++]);
for(j = 0; j < n - 1; j++){
for(max = i, j = i+1; j < n; j++)
if(s[j] > s[max])
max = i;
if(max != j){
p = s[j];
s[j] = s[max];
s[max] = p;
for(j = 0; j < m; j++){
p = a[j][i];
a[j][i] = a[j][max];
a[j][max] = p;
}
}
}
printf("New matrix: \n");
for(i = 0; i < m; i++){
for(j = 0; j < n; printf("%8.2lf", a[i][j++]));
printf("\n");
}
for(j = 0; j < m; j++)
printf("-------------");
printf("\n");
for(j = 0; j < m; printf("%8.2f \n", s[j++]));
printf("\n");
}
}
You can sort the rows of the matrix from highest to lowest, using a simple bubble sort algorithm.Your code modified below:
int main() {
double a[N][N], s[N], p;
int i, j, m, n, max;
while (1) {
printf("\nm, n? ");
scanf("%d%d", & m, & n);
if (m <= 0 || m > N || n <= 0 || n > N)
break;
for (i = 0; i < m; i++) {
printf("%2d. row? ", i + 1);
for (j = 0; j < n; scanf("%lf", & a[i][j++]));
}
for (i = 0; i < m; i++)
for (s[i] = j = 0; j < n; s[i] += a[i][j++]);
for (i = 0; i < m - 1; i++) { // modified here
for (j = i + 1; j < m; j++) { // modified here
if (s[j] > s[i]) { // modified here
p = s[i];
s[i] = s[j];
s[j] = p;
for (int k = 0; k < n; k++) {
p = a[i][k];
a[i][k] = a[j][k];
a[j][k] = p;
}
}
}
}
printf("New matrix: \n");
for (i = 0; i < m; i++) {
for (j = 0; j < n; printf("%8.2lf", a[i][j++]));
printf("\n");
}
for (j = 0; j < m; j++)
printf("-------------");
printf("\n");
for (j = 0; j < m; printf("%8.2f \n", s[j++]));
printf("\n");
}
return 0;
}
Here's how i modified your code to achieve that:
Initialize a loop variable i to 0.
In the outer loop, run the inner loop j from i+1 to m-1.
In the inner loop, compare the sum of the row i with the sum of row
j. If the sum of row j is greater than the sum of row i, swap the
rows using a temporary variable.
After the inner loop finishes, increment the value of i by 1. Repeat
the outer loop until i becomes equal to m-1.
Output:
You can just use qsort to let it handle the sorting and item swapping. Then you only need to write the code for comparing two rows with each other.
Given something like this:
int matrix[3][4] =
{
{1,2,3,4},
{5,6,7,8},
{9,1,2,3},
};
You'd call qsort as:
qsort(matrix, 3, sizeof(int[4]), compare);
The only complexity is implementing the comparison callback function. There's two things to consider there:
We've told qsort that we have an array of 3 items, each of type int[4]. So the void pointers it passes along to us will actually be pointers to type int[4]. That is: int(*)[4].
qsort sorts in ascending order by default, where the item considered "less" ends up first. So we need to tweak that to get the largest item first.
Example:
int compare (const void* obj1, const void* obj2)
{
const int (*ptr1)[4] = obj1;
const int (*ptr2)[4] = obj2;
size_t sum1=0;
size_t sum2=0;
for(size_t i=0; i<4; i++)
{
sum1 += (*ptr1)[i];
sum2 += (*ptr2)[i];
}
if(sum1 > sum2) // largest sum considered "less" for qsort
return -1;
else
return 1;
return 0;
}
sum1 < sum2 would have placed the smallest row first.
Full example:
#include <stdio.h>
#include <stdlib.h>
int compare (const void* obj1, const void* obj2)
{
const int (*ptr1)[4] = obj1;
const int (*ptr2)[4] = obj2;
size_t sum1=0;
size_t sum2=0;
for(size_t i=0; i<4; i++)
{
sum1 += (*ptr1)[i];
sum2 += (*ptr2)[i];
}
if(sum1 > sum2) // largest sum considered "less" for qsort
return -1;
else
return 1;
return 0;
}
void print_matrix(size_t col, size_t row, int matrix[col][row])
{
for(size_t i=0; i<col; i++)
{
for(size_t j=0; j<row; j++)
{
printf("%d,", matrix[i][j]);
}
puts("");
}
}
int main (void)
{
int matrix[3][4] =
{
{1,2,3,4},
{5,6,7,8},
{9,1,2,3},
};
print_matrix(3,4,matrix);
puts("");
qsort(matrix, 3, sizeof(int[4]), compare);
print_matrix(3,4,matrix);
}

How to randomize a to p without repitition

I want to randomize a to p without repetition.
int main(){
int array2[4][4];
bool arr[100]={0};
int i;
int j;
srand(time(NULL));
for(i=0; i<=3; i++){
for(j=0; j<=3; j++){
int randomNumber1;
randomNumber1 = (rand() % (82-65+1))+65;
if (!arr[randomNumber1])
{
printf("%c ",randomNumber1);
array2[i][j]=randomNumber1;
}
else
{
i--;
j--;
arr[randomNumber1]=1;
}
}
printf("\n");
}
return;
the output still has repeat alphabet. I want to have the output in 4x4 with with all a to p without it repeating.
There are some errors in your code. IMHO the most serious is that arr[randomNumber1]=1; is is the wrong branch of the test. That means that your current code does not invalidate once a number was used but only if it has already been invalidated => if you control the arr array at the end of the program all value are still 0.
That is not all. When you get a duplicate, you should only reset the inner loop, and you are currently off by 2 in your maximum ascii code: you go up to R when you want to stop at P.
Your code should be:
for (i = 0; i <= 3; i++) {
for (j = 0; j <= 3; j++) {
int randomNumber1;
randomNumber1 = (rand() % (81 - 65)) + 65;
if (!arr[randomNumber1])
{
printf("%c ", randomNumber1);
array2[i][j] = randomNumber1;
arr[randomNumber1] = 1;
}
else
{
//i--;
j--;
}
}
printf("\n");
}
But this kind of code is terribly inefficient. In my tests it took 30 to 60 steps to fill 16 values, because random can return duplicates. This is the reason why you were advised in comments to use instead the modern algorithm for Fisher-Yates shuffle:
int main() {
int array2[16];
unsigned i, j, k=0;
// initialize array with alphabets from A to P
for (i = 0; i < sizeof(array2); i++) {
array2[i] = 'A' + i;
}
// Use Fisher-Yates shuffle on the array
srand(time(NULL));
for (i = 15; i > 0; i--) {
j = rand() % (i + 1);
if (j != i) {
int c = array2[i];
array2[i] = array2[j];
array2[j] = c;
}
}
// Display a 4x4 pattern
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
printf("%c ", array2[k++]);
}
printf("\n");
}
return 0;
}
Which shuffles the array in only 16 steps.
Here is the outline
// Need some #includes here - exercise for the reader
char items[] = "abcdefghijklmnopqrstuvwxyz";
int len = sizeof(items);
srand(time(NULL));
while (len > 0) {
int r = rand() % len;
printf("%c", items[r]);
len--;
items[r] = items[len];
}
This should do the trick to print the whole alphabet in random order without repeats. Modify to do what you need it to do

Reverse Bubble Sort Algorithm in C using only while loop

I am trying to create a reverse bubble sort algorithm. It works but the first output is a big crazy number that i dont understand. The remaining outputs seem to be sorted in descending order. Where is my code wrong?
#include <stdio.h>
void ft_rev_int_tab(int *tab, int size)
{
int i;
int j;
int k;
i = 0;
while (i < size)
{
j = 0;
while (j < size -i)
{
if (tab[j] < tab[j+1])
{
k = tab[j];
tab[j] = tab[j+1];
tab[j + 1] = k;
}
j++;
}
i++;
}
}
int main(void)
{
int tab[] = {9, 320, 0, 113, 15};
int size = sizeof(tab) / sizeof(*tab);
ft_rev_int_tab(tab, size);
for (int i = 0; i < size; i++)
{
printf ("%d\n", tab[i]);
}
}
i = 0; before while (i < size) is wrong. The condition j < size -i is equivalent to j < size when i is zero. In this case j will be at most size-1 and now tab[j+1] is out-of-range.
It should be changed to i = 1;.
void ft_rev_int_tab(int *tab, int size) {
int i;
int j;
int k;
i = 0;
while (i < size) {
j = 0;
while (j < size - i - 1) {
if (tab[j] < tab[j + 1]) {
k = tab[j];
tab[j] = tab[j + 1];
tab[j + 1] = k;
}
j++;
}
i++;
}
}
Your condition in the nested while loop should be (j < size - i - 1)
because you only need to check the value from tab[0] to tab[i-1]. Your if and swap checks tab[j] and tab[j+1], so when i is equal to the size of the array, you will compare it with a value outside of the array. In your example is the tab[5]'s value.

Print letters in lower triangle in C

I spent hours in printing a lower triangle in C. However, I just cannot figure out how to solve this same question with array.
Below is one of the solution I found on net:
int main(void)
{
char ch='A';
int i,j;
for(i=1;i<7;i++)
{
for(j=0;j<i;j++)
printf("%c",ch++);
printf("\n");
}
return 0;
}
Below is how I try to do the same thing:
#define SIZE 8
int main(void){
char Alphabet[SIZE];
int i, j;
for (i = 0, j = 'A'; i < SIZE, j < 'A' + SIZE; i++, j++){
Alphabet[i] = j;
}
for (i = 0; i <= 7; i++){
for (j = 0; j <= i; j++){
printf("%c", Alphabet[j+i]);
}
printf("\n");
}
return 0;
}
The result of the code above is :
A
BC
CDE
DEFG
EFGHI
FGHIJK
GHIJKLM
HIJKLMNO
What should I revise if I want to print as follow:
A
BC
DEF
GHIJ
KLMNO
PQRSTU
Thank you.
Keep a track of elements printed from the Alphabet array so far and in the inner loop start printing from next element onward. You can do:
#include <stdio.h>
#define SIZE 26
int main(void) {
char Alphabet[SIZE];
for (int i = 0; i < SIZE; i++) {
Alphabet[i] = 'A' + i;
}
// Or simply have the Alphabet array initialized like this
// char Alphabet[SIZE] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
int k = 0;
for (int i = 0; i < 6; i++) {
for (int j = 0; j <= i && k < SIZE; j++) {
printf("%c", Alphabet[k++]);
}
printf("\n");
}
return 0;
}
Output:
# ./a.out
A
BC
DEF
GHIJ
KLMNO
PQRSTU
EDIT:
In the comments, a fellow SO contributor said that the above approach is same as the one OP already found as a solution and OP might be looking for approach of calculating the Alphabet array index using i and j only and without use of variable keeping track of array index. Below is the program which does not use any extra variable to keep the track of Alphabet array index to print characters in inner loop and calculating the index using i and j:
#include <stdio.h>
#define SIZE 26
#define ARRLOC(x) ((x * ((x + 1) / 2)) + ((x % 2 == 0) ? (x / 2) : 0))
int main(void) {
char Alphabet[SIZE];
for (int i = 0; i < SIZE; i++){
Alphabet[i] = 'A' + i;
}
// Or simply have the Alphabet array declared like this
// char Alphabet[SIZE] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int i = 0; i < 6; i++){
for (int j = 0; j <= i && (ARRLOC(i) + j) < SIZE; j++){
printf("%c", Alphabet[ARRLOC(i) + j]);
}
printf("\n");
}
return 0;
}
Output:
# ./a.out
A
BC
DEF
GHIJ
KLMNO
PQRSTU
You can just have a third 'index' variable that keeps track of which letter to output across both loops (I've called this k in the code below). Also, you need to make your Alphabet array bigger (26 seems like a reasonable number); then, if that k variable gets past 'Z', we can simply loop back to 'A' using the modulo operator (%):
#include <stdio.h>
#define SIZE 26
int main(void)
{
char Alphabet[SIZE];
int i, j, k;
for (i = 0; i < SIZE; i++) Alphabet[i] = 'A' + i;
int k = 0;
for (i = 0; i <= 7; i++) {
for (j = 0; j <= i; j++) {
printf("%c", Alphabet[k % 26]); // If past the end, loop back with the "%" operator
++k;
}
printf("\n");
}
return 0;
}
Or we can make the code a little more 'succinct' (though perhaps less clear) by initializing the k variable at the start of the outer loop and incrementing at the end of the inner loop:
for (k = i = 0; i <= 7; i++) { // Initialize "k" here ...
for (j = 0; j <= i; j++, k++) { // .. but increment it here!
printf("%c", Alphabet[k % 26]); // If past the end, loop back with the "%" operator
}
printf("\n");
}
If you want an 8 by 8 pyramid, you won't have enough characters to do it using the alphabet (requires 36), so I made the alphabet repeat itself (u could also make it go to numeric instead?)
#define SIZE 8
int area(int size);
int main(void){
char Alphabet[area(SIZE)];
int i, j;
for (i = 0, j = 'A'; i < area(SIZE); i++, j++){
if (j > 'Z') j = 'A';
Alphabet[i] = j;
}
int idx=0;
for (i = 0; i < SIZE; i++){
for (j = 0; j <= i; j++){
printf("%c", Alphabet[idx++]);
}
printf("\n");
}
return 0;
}
int area(int size) {
if (size==1) return 1;
return size + area(size - 1);
}

Preventing duplicates from being stored when filling dynamic array with values from two different arrays

I need to know a method to keep duplicate numbers from being stored in a new array when taking numbers from two different arrays. The function is supposed to store each 'unique' value once and not store duplicate values again.
Here is my function code so far:
int * arrayIntersect(int *sizeOfResult, const int *a, const int *b, int sizeOfA, int sizeOfB){
int i;
int j;
int k = 0;
int c[(sizeOfA + sizeOfB)];
for(j = 0; j < sizeOfB; j++){
for(i = 0; i < sizeOfA; i++){
if(a[i] == b[j]){
c[k] = a[i];
(*sizeOfResult)++;
k++;
}
}
}
int *d = (int *)malloc(sizeof(int) * *sizeOfResult);
for(i = 0; i < *sizeOfResult; i++){
d[i] = c[i];
}
return d;
}
It prints the values I need, but I want to eliminate the same number from showing up multiple times when printing the contents of the new dynamic array.
Any idea on how to improve my code to allow prevent duplication?
The proper way to do it is having the arrays ordered and then doing a binary search for each insertion like #Murilo Vasoncelos pointed out.
Below is a quick and dirty solution that loops through a and b and for each iteration checks if the number has been inserted before. If it isn't, it inserts it.
int duplicate = 0;
*sizeOfResult = 0;
for(j = 0; j < sizeOfA; j++){
for(i = 0; i < (*sizeOfResult); i++){
if(c[i] == a[j]){
duplicate = 1;
break;
}
}
if (!duplicate)
{
c[(*sizeOfResult)] = a[i];
(*sizeOfResult)++;
}
duplicate = 0;
}
for(j = 0; j < sizeOfB; j++){
for(i = 0; i < (*sizeOfResult); i++){
if(c[i] == b[j]){
duplicate = 1;
break;
}
}
if (!duplicate)
{
c[(*sizeOfResult)] = b[i];
(*sizeOfResult)++;
}
duplicate = 0;
}
If your arrays a and b are ordered, you can simply use this linear algorithm for array intersection:
int* inter(int* szr, int* a, int* b, int sza, int szb)
{
int c[MAX(sza, szb)];
int i, j, k = 0;
for (i = 0, j = 0; i < sza && j < szb;) {
if (a[i] == b[j]) {
if (k == 0 || c[k - 1] < a[i]) {
c[k++] = a[i];
}
i++;
j++;
}
else if (a[i] < b[j]) {
i++;
}
else {
j++;
}
}
*szr = k;
int* ans = (int*)malloc(sizeof(int) * k);
for (i = 0; i < k; ++i) {
ans[i] = c[i];
}
return ans;
}

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