# The question asks to find the maximum and minimum spacing after clustering
# the answer lies in the remaining edges in the heap.
# First build a node dictionary for each node pointing to what cluster it is in
# Also build a cluster dictionary that stores one value for each of N to N clusters
# Then keep popping from the heap, and store the edge cost in the cluster dictionary according to which cluster each
# of it's nodes are in
# Once the cluster dictionary is full, the largest value is the maximum spacing, and the smallest is the minimum
# Creating the node dictionary runs in O(n)
# Creating the cluster dictionary runs in less then O(k^2).
# Therefore if k is sufficiently smaller then n, this doesn't effect the overall running time of O(m|alpha(N))
node_to_cluster = {}
c = 0
for cluster in union_find.get_components():
for node in cluster:
node_to_cluster[node] = str(c)
c+=1
cluster_dict = {}
for i in range(k):
for j in range(i, k):
if j > i:
cluster_1 = str(i)
cluster_2 = str(j)
cluster_dict[cluster_1+"_"+cluster_2] = None
while None in cluster_dict.values():
(cost, (node_1, node_2)) = heapq.heappop(edge_heap)
