Count Triplets That Can Form Two Arrays of Equal XOR - Problem

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* arr, int n) {
    if (n < 2) return 0;
    
    // Build prefix XOR array
    int* prefixXor = (int*)calloc(n + 1, sizeof(int));
    for (int i = 0; i < n; i++) {
        prefixXor[i + 1] = prefixXor[i] ^ arr[i];
    }
    
    int count = 0;
    // For each starting position i
    for (int i = 0; i < n; i++) {
        // For each ending position k
        for (int k = i + 1; k < n; k++) {
            // If XOR from i to k is 0, then we can split it
            if ((prefixXor[k + 1] ^ prefixXor[i]) == 0) {
                // Any j between i+1 and k (inclusive) creates a valid triplet
                count += k - i;
            }
        }
    }
    
    free(prefixXor);
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int arr[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        arr[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(arr, n);
    printf("%d\n", result);
    return 0;
}