This is an answer to this problem.
Basically, given an array, swap the given number of elements in the array.
The solution that came to my mind: since we always swap at least two elements, pick two random indices and swap array[first_random] with array[second_random]. Then, if there's more to swap, find index that was not swapped yet and swap it with one of previously swapped.
This might be confusing a little, but it works. I am curious whether there are some other, better approaches.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void print_arr(int a[], int size) {
int i;
for (i = 0; i < size; i++) {
printf("%d ",a[i]);
}
printf("\n");
}
void swap(int a[], int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
int next_idx(int swapped[], int s_count, int size) {
int n, i;
char in_arr;
while (1) {
in_arr = 0;
n = rand() % size;
for (i = 0; i < s_count; i++) {
if (swapped[i] == n) {
in_arr = 1;
break;
}
}
if (!in_arr) {
break;
}
}
return n;
}
void swap2_or_more(int a[], int size,int count) {
srand(time(NULL));
int i, j, s_count = 0;
int swapped[size];
i = rand() % size;
swapped[s_count] = i;
s_count++;
do {
j = rand() % size;
} while (i == j); // make sure indexes are different
swapped[s_count] = j;
s_count++;
swap(a, i, j);
count -= 2;
while (count) {
i = next_idx(swapped, s_count, size);
j = rand() % s_count;
swap(a, i, swapped[j]);
swapped[s_count] = i;
s_count++;
count--;
}
printf("swapped indexes:\n");
print_arr(swapped, s_count);
}
int main(void)
{
int a[] = {1,2,3,4,5,6,7,8,9,10};
swap2_or_more(a, 10, 5);
printf("array after swap:\n");
print_arr(a, 10);
return 0;
}