Consequences of disjoint set union path compression in union by size

Question

When doing union by size, you compare the size of the nodes that you're trying to unify. I found this code on codeforces that shows how to build a DSU with path compression and union by size:

#include <bits/stdc++.h>
using namespace std;

class DSU
{
public:
    unordered_map<int, int> parent;
    unordered_map<int, int> size;

    void make_set(int v)
    {
        parent[v] = v;
        size[v] = 1;
    }

    int find_set(int v)
    {
        if (v == parent[v])
            return v;

        return parent[v] = find_set(parent[v]);
    }

    void union_sets(int a, int b)
    {
        a = find_set(a);
        b = find_set(b);

        if (a != b)
        {
            if (size[a] < size[b])
                swap(a, b);

            // a big, b small
            // attach small tree to big tree
            parent[b] = a;
            size[a] += size[b];
        }
    }
};

The question is, don't you have to modify find_set to decrease the size of the parent node since the parent node will be pointing to the root now and not to another child node ?

Wouldn't this be the correct implementation of find_set?

    int find_set(int v)
    {
        if (v == parent[v])
            return v;

        size[parent[v]] -= 1;
        return parent[v] = find_set(parent[v]);
    }

Answer 1

Yes, but you gain nothing since that parent is not the representative of the set.

That is,

int find_set(int v)
{
    if (v == parent[v])
        return v;
    
    // This parent is not the representative, changing
    // the size is superfluous as it will no longer 
    // be referenced again.
    size[parent[v]] -= 1;
    return parent[v] = find_set(parent[v]);
}