Inversion Count -Merge Sort(Efiicient)

   long long int countAndMerge(long long arr[], long long l, long long m, long long r) {
    long long int n1 = m - l + 1;
    long long int n2 = r - m;
    vector<long long int> left(n1);
    vector<long long int> right(n2);

    for (int i = 0; i < n1; i++) {
        left[i] = arr[l + i];
    }
    for (int i = 0; i < n2; i++) {
        right[i] = arr[m + 1 + i];
    }

    long long int i = 0, j = 0, k = l;
    long long int res = 0;

    while (i < n1 && j < n2) {
        if (left[i] <= right[j]) {
            arr[k++] = left[i++];
        } else {
            arr[k++] = right[j++];
            res += (n1 - i); // Counting the inversions
        }
    }

    while (i < n1) {
        arr[k++] = left[i++];
    }
    while (j < n2) {
        arr[k++] = right[j++];
    }

    return res;
}

long long int merge(long long arr[], long long l, long long r) {
    long long int p = 0;
    if (l < r) {
        long long int m = l + (r - l) / 2;

        // Recursively count inversions in left and right subarrays
        p += merge(arr, l, m);
        p += merge(arr, m + 1, r);

        // Count inversions during the merge process
        p += countAndMerge(arr, l, m, r);
    }
    return p;
}

long long int inversionCount(long long arr[], int n) {
    return merge(arr, 0, n - 1);
}