Thanks for sharing, this is very interesting. Note that the status one step before completion must have exactly one node without any connections, with other nodes fully connected. Probably we can enumerate this node (n possibilities). With this node fixed, the other nodes to flip seem fixed as well to me, and to determine each possibility would be O(n) time, so overall TC is O(n^2). Might do some coding later when I have time.

Edit: I feel this brain-teaser style problem amazing so I wrote the full code with test cases :) feel free to correct me if anything is wrong.

```
from typing import Optional
def complete_day(graph: list[list[int]]) -> Optional[set[int]]:
'''
The status before completion must have one node without any connections,
with other nodes fully connected. We can enumerate this node.
Parameter: n*n matrix, graph[i][j] = graph[j][i] = 1
if there is an edge between i and j, else 0.
Return: subset of indexes to flip. If such flip does not exist, return None.
'''
n = len(graph)
# Preprocess: count initial edges for each node
init_edges = [sum(edges) for edges in graph]
if all(edges == n for edges in init_edges):
# Already complete
return set()
# Enumerate the node to leave out at the last step
for i in range(n):
# We have to flip all the nodes with an edge to i,
# and leave others intact.
to_flip = {j for j in range(n) if graph[i][j]}
# flip everything == flip nothing
if len(to_flip) == n:
continue
# Check if this is a legit flip
for j in range(n):
if j in to_flip:
if init_edges[j] != 2 and i != j:
# Only edges between j-j and j-i allowed
break
else:
if init_edges[j] != n - len(to_flip):
# Should be fully connected, except with to_flip subset
break
else:
# The for loop is completed without a break,
# so we already found the subset
return to_flip
# No such flip exists
return None
# Some test cases
# {0,1} - Example 1 (identical to flip {2})
print(complete_day([
[1,1,0],
[1,1,0],
[0,0,1],
]))
# {0,1} - Example 2
print(complete_day([
[1,1,0,0],
[1,1,0,0],
[0,0,1,1],
[0,0,1,1],
]))
# None - Example 3
print(complete_day([
[1,1,0],
[1,1,1],
[0,1,1],
]))
# empty set
print(complete_day([
[1,1,1],
[1,1,1],
[1,1,1],
]))
# {2,3,4}
print(complete_day([
[1,1,0,0,0],
[1,1,0,0,0],
[0,0,1,0,1],
[0,0,0,1,1],
[0,0,1,1,1],
]))
# {0,1}
print(complete_day([
[1,1,0,0,0],
[1,1,0,0,0],
[0,0,1,1,1],
[0,0,1,1,1],
[0,0,1,1,1],
]))
# None
print(complete_day([
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,1,1,1],
[0,0,1,1,1],
[0,0,1,1,1],
]))
```