Union-Find Algorithm | Set 2 (Union By Rank and Path Compression) | GeeksforGeeks

The idea is to always attach smaller depth tree under the root of the deeper tree. This technique is called union by rank.

The second optimization to naive method is Path Compression. The idea is to flatten the tree whenfind() is called. When find() is called for an element x, root of the tree is returned. The find() operation traverses up from x to find root. The idea of path compression is to make the found root as parent of x so that we don’t have to traverse all intermediate nodes again. If x is root of a subtree, then path (to root) from all nodes under x also compresses.

The time complexity of each operations becomes even smaller than O(Logn). In fact, amortized time complexity effectively becomes small constant.

struct subset
{
    int parent;
    int rank;
};
int find(struct subset subsets[], int i)
{
    // find root and make root as parent of i (path compression)
    if (subsets[i].parent != i)
        subsets[i].parent = find(subsets, subsets[i].parent);
 
    return subsets[i].parent;
}
void Union(struct subset subsets[], int x, int y)
{
    int xroot = find(subsets, x);
    int yroot = find(subsets, y);
 
    // Attach smaller rank tree under root of high rank tree
    // (Union by Rank)
    if (subsets[xroot].rank < subsets[yroot].rank)
        subsets[xroot].parent = yroot;
    else if (subsets[xroot].rank > subsets[yroot].rank)
        subsets[yroot].parent = xroot;
 
    // If ranks are same, then make one as root and increment
    // its rank by one
    else
    {
        subsets[yroot].parent = xroot;
        subsets[xroot].rank++;
    }
}
 
// The main function to check whether a given graph contains cycle or not
int isCycle( struct Graph* graph )
{
    int V = graph->V;
    int E = graph->E;
 
    // Allocate memory for creating V sets
    struct subset *subsets =
        (struct subset*) malloc( V * sizeof(struct subset) );
 
    for (int v = 0; v < V; ++v)
    {
        subsets[v].parent = v;
        subsets[v].rank = 0;
    }
 
    // Iterate through all edges of graph, find sets of both
    // vertices of every edge, if sets are same, then there is
    // cycle in graph.
    for(int e = 0; e < E; ++e)
    {
        int x = find(subsets, graph->edge[e].src);
        int y = find(subsets, graph->edge[e].dest);
 
        if (x == y)
            return 1;
 
        Union(subsets, x, y);
    }
    return 0;
}

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