How to Avoid repeating number randomized in C - c

Question

int i,j,vec[15]={0};
srand (time(NULL));
for (i=0;i<15;i++){
vec[i]=rand() % 25+1;
for (j=0;j<15;j++){
if (i!=j){
while(vec[i]==vec[j]){
vec[i]=rand() % 25+1;
}
}
}
printf("%d\n",vec[i]);
}
return 0;
}
the code still gives me repeated numbers
EXAMPLE:
24
3
7
20
18
10
12
17
9
7
4
25
13
15
21
I cant figure out what to do with it

Answer 1

You have your loops mixed up. The logic is: Generate a random number until you have found one that isn't in the list.
The way you do it, you generate a new number inside the checking loop. But that doesn't work. Say you're generating the 4th number and find it is equal to the third. Then you generate a new one which might well be equal to any you have already checked against.
You also check uninitialised elements when j > i. Your inner loop should only run up to i.
So:
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
int main()
{
int vec[15] = { 0 };
int i, j;
srand(time(NULL));
for (i = 0; i < 15; i++) {
int okay = 0;
while (!okay) {
vec[i] = rand() % 25 + 1;
okay = 1;
for (j = 0; j < i; j++) {
if (vec[i] == vec[j]) okay = 0;
}
}
printf("%d\n", vec[i]);
}
return 0;
}
That still looks a bit awkward with that okay variable. In my opinion, checking for duplicates should be a separate function:
int contains(int arr[], int n, int x)
{
while (n--) {
if (arr[n] == x) return 1;
}
return 0;
}
int main()
{
// snip ...
for (i = 0; i < 15; i++) {
do {
vec[i] = rand() % 25 + 1;
} while (contains(vec, i, vec[i]));
printf("%d\n", vec[i]);
}
// snip ...
}
In your case the range of possible numbers isn't mich bigger than the number of array elements. You could also create an ordered array {1, 2, 3, ..., 25}, then shuffle it and use only the first 15 elements.

Answer 2

Reset j in the while loop:
for (j=0;j<i;j++){ //Use j<i
if (i!=j){
while(vec[i]==vec[j]){
vec[i]=rand() % 25+1;
j=-1;//-1 because in the next iteration,j will start from 0
}
}
}

Answer 3

Are you actually trying to shuffle the numbers, rather than fill the array with randoms? (It looks like you want an array with numbers from 1 to 25, but in random order.) rand() can give you duplicate numbers (they're random, after all!)
Try this:
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <time.h>
int
main( int argc, char **argv )
{
int i, vec[25];
for (i = 0; i < 25; ++i) vec[i] = i + 1;
/* Shuffle entries */
srand( time( 0 ) );
for (i = 0; i < 1000; ++i) {
int a = rand( ) % 25;
int b = rand( ) % 25;
if (a != b) {
int tmp = vec[a];
vec[a] = vec[b];
vec[b] = tmp;
}
}
/* Print shuffled array */
for (i = 0; i < 25; ++i) printf( "%d: %d\n", i, vec[i] );
return 0;
}

Answer 4

#include<stdio.h>
#include<stdlib.h>
int inArray(int, int, int*);
int main()
{
int i,j,vec[15]={0};
int temp;
srand (time(NULL));
for (i=0;i<15;i++){
temp =rand() % 25+1;
while(inArray(i+1,temp, vec) == 1){
temp = rand() % 25+1;
}
vec[i] = temp;
printf("VECT[%d] \t= %d\n",i,vec[i]);
}
return 0;
}
int inArray(int count, int input, int* array){
int i = 0;
for(i=0; i<count; i++){
if(input == array[i]){
return 1;
}
}
return 0;
}
Gave an output:
VECT[0] = 24
VECT[1] = 19
VECT[2] = 1
VECT[3] = 25
VECT[4] = 22
VECT[5] = 18
VECT[6] = 7
VECT[7] = 8
VECT[8] = 12
VECT[9] = 21
VECT[10] = 11
VECT[11] = 6
VECT[12] = 23
VECT[13] = 20
VECT[14] = 15
The checking was off, you would change and not break allowing it to be changed to a previous value.

Answer 5

You can use an array
int randNumbers[25]; // fill it starting 0 to 25 then
randomize the number in a range between 0 and 25 after swap the number in the randomized index with the last number in your array
randomize 0 to 23
and so on....
int main(int argc, char **argv) {
static const int size = 25;
int numbers[size];
for( int i = 0; i < size; i++ ){
numbers[i] = i;
}
srand (time(NULL));
for( int i = 0; i < size; i++ ){
int rIndex = rand()%(size - i);
int rNum = numbers[rIndex];
numbers[rIndex] = numbers[size-i];
printf("%d ", rNum);
}
return 0;
}
O(n) complexity...