Shuffling array elements in C

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 
#include 
#include 

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;
}

There is a huge amount of guessing in your program, code like:

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;
}

Let's say the input data has 1000000 members, and we need to swap 10 of them... swapped[] array contains.... {1, 100, 500, 800, 1000, 10000} do we really just keep guessing until we hit a random value that is in the data? This is... a problem.

One solution which will produce a suitably random distribution of the swapped items relies on a 'simple' algorithm which does just one rand() per item, and there's no guess-work. This is done using a fisher-yates shuffle. But, for this problem, we don't shuffle the items, we shuffle the indexes... so:

void swap2_or_more(int a[], int size,int count) {
    int i, tmp;
    int swapped[size];

    // initialize the swapped array with all the data indices.
    for (i = 0; i  0; i--) {
        tmp = rand() % (i + 1);
        swap(swapped, i, tmp);
    }

    // 'rotate' the values in the first count swapped/random indexes
    tmp = a[swapped[--count]];
    while(count > 0) {
        a[swapped[count]] = a[swapped[count - 1]];
        count--;
    }
    a[swapped[0]] = tmp;
}

StackExchange Code Review Q#58837, answer score: 4