fastest algorithm count number of 3 length AP in array

Question

I want to solve this CodeChef challenge:

Suppose We are given an array A of N(of range 100,000) elements. We are to find the count of all pairs of 3 such elements 1&

Answer 1

Here's a simple C version of the solution that takes advantage of the Ai + Ak must be even test:

#include 

static int arr[] = {9, 4, 2, 3, 6, 10, 3, 3, 10};

int main ()
{
    int i, j, k;
    int sz = sizeof(arr)/sizeof(arr[0]);
    int count = 0;

    for (i = 0; i < sz - 2; i++)
    {
        for (k = i + 2; k < sz; k++)
        {
            int ik = arr[i] + arr[k];
            int ikdb2 = ik / 2;

            if ((ikdb2 * 2) == ik) // if ik is even
            {
                for (j = i + 1; j < k; j++)
                {
                    if (arr[j] == ikdb2)
                    {
                        count += 1;
                        printf("{%d, %d, %d}\n", arr[i], arr[j], arr[k]);
                    }
                }
            }
        }
    }
    printf("Count is: %d\n", count);
}

and the console dribble:

tmp e$ cc -o triples triples.c
tmp e$ ./triples
{9, 6, 3}
{9, 6, 3}
{2, 6, 10}
{2, 6, 10}
{3, 3, 3}
Count is: 5
tmp e$ 

This more complicated version keeps a list of Aj indexed by value to go from n-cubed to n-squared (kinda).

#include 
#include 

static uint32_t arr[] = {9, 4, 2, 3, 6, 10, 3, 3, 10};

#define MAX_VALUE 100000u
#define MAX_ASIZE  30000u

static uint16_t index[MAX_VALUE+1];
static uint16_t list[MAX_ASIZE+1];

static inline void remove_from_index (int subscript)
{
    list[subscript] = 0u; // it is guaranteed to be the last element

    uint32_t value = arr[subscript];

    if (value <= MAX_VALUE && subscript == index[value])
    {
        index[value] = 0u; // list now empty
    }
}

static inline void add_to_index (int subscript)
{
    uint32_t value = arr[subscript];

    if (value <= MAX_VALUE)
    {
        list[subscript] = index[value]; // cons
        index[value] = subscript;
    }
}

int main ()
{
    int i, k;
    int sz = sizeof(arr)/sizeof(arr[0]);
    int count = 0;

    for (i = 0; i < sz - 2; i++)
    {
        for (k = i; k < sz; k++) remove_from_index(k);

        for (k = i + 2; k < sz; k++)
        {
            uint32_t ik = arr[i] + arr[k];
            uint32_t ikdb2 = ik / 2;

            add_to_index(k-1); // A(k-1) is now a legal middle value 

            if ((ikdb2 * 2) == ik) // if ik is even
            {
                uint16_t rover = index[ikdb2];

                while (rover != 0u)
                {
                    count += 1;
                    printf("{%d, %d, %d}\n", arr[i], arr[rover], arr[k]);

                    rover = list[rover];
                }
            }
        }
    }
    printf("Count is: %d\n", count);
}

and the dribble:

tmp e$ cc -o triples triples.c
tmp e$ ./triples
{9, 6, 3}
{9, 6, 3}
{2, 6, 10}
{2, 6, 10}
{3, 3, 3}
Count is: 5
tmp e$