Count the number of ways to obtain a sum, with consecutivity restriction

A cricketer can score 1, 2, 4 or 6 in a ball. Find in how many ways the player can score a total of "n" runs without hitting 3 consecutive boundaries. Note: Scoring 4 or 6 is considered as a boundary.

My solution is:

int *arr; //global array to save calculated ways
int waysUtil(int n, int boundaries){

     if(n==0)
        return 1;
     if(n0)
        return arr[n];//if already calculated return the value
    int hit_one = waysUtil(n-1, 0);
    if(n-1 >=0) arr[n-1]= hit_one;
    int hit_two = waysUtil(n-2, 0);
    if(n-2 >=0) arr[n-2] = hit_two;
    int hit_four=0, hit_six=0;
    // for 4 and 6 check number of consecutive boundaries hit.
    //if less than two then calculate
    if(boundaries=0) arr[n-4] = hit_four;
        hit_six = waysUtil(n-6, boundaries + 1);
        if(n-6 >=0) arr[n-6] = hit_six;

    }
    return hit_one + hit_two + hit_four + hit_six;
}

int ways(int score){
    arr =(int*)malloc(sizeof(int)*score);
    memset(arr, -1, sizeof(arr));
    int noOfWays = waysUtil(score, 0);
    free(arr);
    return noOfWays;
}

Is better solution possible?

This looks wrong:

memset(arr, -1, sizeof(arr));

This is setting sizeof(int*) bytes of arr to -1. It doesn't set the entire contents of arr.

You should probably be doing this:

int arr_size = sizeof(int)*score;
arr = (int*)malloc(arr_size);
memset(arr, -1, arr_size);

Although as pointed out by @Andrea Biondo, this only works because most architectures and compilers are based around twos complement numbers. There are some exceptions to this rule, which you're likely to know about if they effect you.

StackExchange Code Review Q#134242, answer score: 4