Adding minimum nodes to a disconnected graph in order to be able to reach every node from other, with connectivity constraints

There is this graph where each node has a spatial position (x,y), and the edges between the nodes are only connected if the nodes are on the same vertical or horizontal line. By other words, nodes A (with spatial position(xA, yA)) and B (with spatial position(xB, yB)) are connected if and only if (xA == xB) OR (yA == yB).
There are also some other nodes in the graph which are not connected, because there is no node with the same x or y.
Example:

import networkx as nx

G=nx.Graph()

G.add_node(1,pos=(1,1))
G.add_node(2,pos=(1,2))
G.add_node(3,pos=(1,3))
G.add_node(4,pos=(1,4))
G.add_node(5,pos=(1,5))

G.add_node(6,pos=(2,1))
G.add_node(7,pos=(3,1))
G.add_node(8,pos=(4,1))

G.add_node(9,pos=(5,3))
G.add_node(10,pos=(6,7))
G.add_node(11,pos=(10,6))


G.add_edge(1,2)
G.add_edge(2,3)
G.add_edge(3,4)
G.add_edge(4,5)

G.add_edge(1,6)
G.add_edge(6,7)
G.add_edge(7,8)

G.add_edge(3,9)


pos=nx.get_node_attributes(G,'pos')

nx.draw(G,pos)

Example

Can anyone help me find the algorithm which gives me least number of nodes that should be added so that the graph will be connected (every node can be reached from any other)?

I found this question similar to mine but I couldn't get the algorithm.

Notice that if two unconnected node lie on the same row or column, then they can be connected to the other connected nodes by just one new node.

So:

• Construct a 3D vector [row/column][row or column index][node index]
• LOOP over unconnected nodes
  -Add node index to 3d vector

example: unconnected node at row 3 column 2 you would add its index to vector[row][3] and vector[col][2]

• FIND maximum count of vector[row][row range] AND vector[col][col range]
• Add node that connects the nodes in the found maximum
• Remove newly connected nodes from vector
• Repeat until all nodes connected

Let me know if some C++ code that implements this would be useful.

If we run DFS over the graph and count each time the DFS ran, the result shows us the number of connected components in the graph. We need one node for each new component (to connect it to the rest). So the result would be the counted number of components - 1!