(This question has been seen in the interviews of the following companies: Google, Facebook)

##### [Google Onsite Questions]

# Set Matrix Zeroes II

There are orbs on an NxM board ('O' denotes orb. 'X' denotes empty slot).

Whenever two orbs are in the same column or the same row, you get to erase one of them.

Try to erase as many orbs as possible.

Find the minimum number of orbs remained on board eventually.

e.g.OXOXXO XXXXXO XXXXOXerase (0,0) and get

XXOXXO XXXXXO XXXXOXerase (2,0) and get

XXXXXO XXXXXO XXXXOXerase (5,0) and get

XXXXXX XXXXXO XXXXOXno more moves available. Return 2 e.g.

OXOXXO XXXXXO XXOXOX

erase(4,2), (2,2), (0,0), (2,0), (5,0)

Return 1

e.g.

OXOXXX XOXXXO XXXOOX

erase(0,0), (1,1), (3,2)

Return 3

**Follow-up**
Build a solver for this board game that erases the as many orbs as possible.

## Implementation

```
def solve(board):
m, n = len(board), len(board[0])
'''create adjacency list
rows[0] has all the nodes in 1st row
rows[1] has all the nodes in 2nd row
etc.
cols[0] has all the nodes in 1st column
cols[1] has all the nodes in 2nd column
etc.'''
rows, cols = [set() for i in range(m)], [set() for j in range(n)]
for i in range(m):
for j in range(n):
if board[i][j] is 'O':
rows[i].add(j)
cols[j].add(i)
#count the degree (number of neighbors) of every orb
degrees = dict()
for i in range(m):
for j in rows[i]:
degree = len(rows[i]) + len(cols[j]) - 2
if degree > 0:
degrees[(i, j)] = degrees
#erase the node with minimum positive degree until all nodes have 0 degree
while degrees:
erase_orb(degrees, board, rows, cols)
#output
for row in board:
print row
#erase the orb with the minimum degree
def erase_orb(degrees, board, rows, cols):
#find the node with minimum positive degree
i, j = min(degrees.items(), key=lambda x: x[1])[0]
#erase this node
rows[i].remove(j)
cols[j].remove(i)
degrees.pop((i, j))
board[i][j] = 'X'
print 'erase ', (i, j)
#remove one degree from every neighbor of the node erased
for y in rows[i]:
degrees[(i, y)] -= 1
if degrees[(i, y)] is 0:
degrees.pop((i, y))
for x in cols[j]:
degrees[(x, j)] -= 1
if degrees[(x, j)] is 0:
degrees.pop((x, j))
solve([
['X', 'X', 'X', 'X', 'X', 'O'],
['O', 'X', 'X', 'X', 'X', 'X'],
['O', 'X', 'X', 'X', 'X', 'O'],
['X', 'X', 'O', 'X', 'O', 'X'],
['X', 'O', 'X', 'X', 'X', 'O']
])
```

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