// Union-Find Disjoint Sets Library written in OOP manner, using both path compression and union by rank heuristics
class UnionFind {                                              // OOP style
private:
  vi p, rank, setSize;                       // remember: vi is vector<int>
  int numSets;
public:
  UnionFind(int N) {
    setSize.assign(N, 1); 
    numSets = N; 
    rank.assign(N, 0);
    p.assign(N, 0); for (int i = 0; i < N; i++) p[i] = i; }
  
  int findSet(int i) { return (p[i] == i) ? i : (p[i] = findSet(p[i])); }
  
  bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }
  
  void unionSet(int i, int j) { 
    if (!isSameSet(i, j)) { 
      numSets--; 
      int x = findSet(i), y = findSet(j);
      // rank is used to keep the tree short
      if (rank[x] > rank[y]) { 
        p[y] = x; 
        setSize[x] += setSize[y]; }
      else { 
        p[x] = y; 
        setSize[y] += setSize[x];
        if (rank[x] == rank[y]) 
          rank[y]++; }}}

  int numDisjointSets() { return numSets; }
  int sizeOfSet(int i) { return setSize[findSet(i)]; }
};

What I could infer: find funtion gives the key of a particular set. we are setting p[i]=find(p[i]), this will set the key of the set as parent of every element of the set.

Doubt: Some elements will still have p[i]=nk where nk is an element of the set which is not the key. If this is the case ,then why perform this assignment (p[i]=find(p[i]))