Week 2 Scrabble Optimization

I just finished the Scrabble problem from Week 2. Frankly, it took me about a day to wrap my head around it. I think what I did was very unorthodox and unoptimized at all. Can you help me find ways to optimize this code? I know programmers have relatively different solutions for solving things. Need your guys insights! Thank you in advance! Here's mine:

#include <ctype.h>
#include <cs50.h>
#include <stdio.h>
#include <string.h>

// Points assigned to each letter of the alphabet
int POINTS[] = {1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 1, 1, 1, 4, 4, 8, 4, 10};
char ALPHABET[] = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};

int compute_score(string word);
const int N = 2;

int main(void)
{
    // Get input words from players
    string playerWord[N];
    int playerNumber = 1;
    for (int i = 0; i < N; i++)
    {
        playerWord[i] = get_string("Player %i: ", playerNumber);
        playerNumber++;
    }
    // Score both words
    int score[N];
    score[0] = compute_score(playerWord[0]);
    score[1] = compute_score(playerWord[1]);

    // TODO: Print the winner

    if (score[0] > score[1])
    {
        printf("Player 1 wins!\n");
    }
    else if (score[1] > score[0])
    {
        printf("Player 2 wins!\n");
    }
    else
    {
        printf("Tie!\n");
    }
}


int compute_score(string word)
{
    // TODO: Compute and return score for string
    int n = strlen(word);
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        int counter = 0;
        word[i] = toupper(word[i]);
        for (int j = 0; j < 27; j++)
        {
            if (word[i] >= 'A' && word[i] <= 'Z')
            {
                if (word[i] == ALPHABET[j])
                {
                    sum += POINTS[counter];
                    break;
                }
                counter++;
            }
            else
            {
                break;
            }
        }
    }
    return sum;
}

​