The answer seems wrong.

    long long int totalWays = n*(n-1) / 2;
    long long int sameWays = 0;
    
    for(i=0;i<numComponents;i++)
    {    
        sameWays = sameWays + (eachC[i]*(eachC[i]-1) / 2);
    }
    cout << (totalWays - sameWays) << endl;
    return 0;

after finding components and number on each component, with the same syntax used in the editorial soln, the final calculation should be:

    int the_count = 0, total=0;
    for (i=0;i<numComponents;i++) {
        the_count += eachC[i] * total;
        total += eachC[i];
    }
    return the_count;

This is basically keeping the property of the_count: it is the number of possible interactions until ith component. Addition of each new component creates itself times the total number of nodes; as an addition to previous possibilities.

This changes half of the test cases.