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Program to count number of walls required to partition top-left and bottom-right cells in Python
Suppose we have a 2d binary matrix where 0 represents empty cell and 1 represents a wall. We have to find the minimum number cells that need to become walls so that there will be no path between top−left cell and bottom-right cell. We cannot put walls on the top−left cell and the bottom−right cell. We can move only left, right, up and down not diagonally.
So, if the input is like
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 0 | 0 | 0 | 0 |
then the output will be 2,
| 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 0 | 0 | 1 | 0 |
To solve this, we will follow these steps −
R := row count of matrix, C := column count of matrix
visited := a new set
tin := a new map, low := a new map
timer := 0
bridge_pts := a new set
par := a new map
src := a pair (0, 0)
tgt := a pair (R − 1, C − 1)
Define a function dfs() . This will take v, parent
mark v as visited
par[v] := parent, tin[v] := timer, low[v] := timer
timer := timer + 1
children := 0
for each to neighbors of v, do
if to is same as parent, then
go for next iteration
if to is visited, then
low[v] := minimum of low[v] and tin[to]
otherwise,
dfs(to, v)
low[v] := minimum of low[v] and low[to]
if low[to] >= tin[v] and parent is not null, then
add v into bridge_pts
children := children + 1
if parent is null and children > 1, then
add v into bridge_pts
Define a function bfs() . This will take root
Q := a double ended queue with a list with single element root
visited := a new set and initially insert root
while Q is not empty, do
v := last element of Q, then delete last element from Q
if v is same as tgt, then
return True
for each w in the neighbors of v, do
if w is not visited, then
mark w as visited
insert w at the left of Q
return False
From the main method do the following −
dfs(src, null)
if tgt is not in par, then
return 0
for each pair (i, j) in bridge_pts, do
matrix[i, j] := 1
if bfs(src) is true, then
return 2
return 1
Let us see the following implementation to get better understanding −
Example
from collections import deque
class Solution:
def solve(self, matrix):
R = len(matrix)
C = len(matrix[0])
def get_neighbors(i, j):
for ii, jj in ((i + 1, j), (i− 1, j), (i, j + 1), (i, j − 1)):
if 0 <= ii < R and 0 <= jj < C and matrix[ii][jj] == 0:
yield ii, jj
visited = set()
tin = {}
low = {}
timer = 0
bridge_pts = set()
par = {}
src = (0, 0)
tgt = (R− 1, C− 1)
def dfs(v, parent):
nonlocal timer
visited.add(v)
par[v] = parent
tin[v] = timer
low[v] = timer
timer += 1
children = 0
for to in get_neighbors(*v):
if to == parent:
continue
if to in visited:
low[v] = min(low[v], tin[to])
else:
dfs(to, v)
low[v] = min(low[v], low[to])
if low[to] >= tin[v] and parent is not None:
bridge_pts.add(v)
children += 1
if parent is None and children > 1:
bridge_pts.add(v)
def bfs(root):
Q = deque([root])
visited = set([root])
while Q:
v = Q.pop()
if v == tgt:
return True
for w in get_neighbors(*v):
if w not in visited:
visited.add(w)
Q.appendleft(w)
return False
dfs(src, None)
if tgt not in par:
return 0
for i, j in bridge_pts:
matrix[i][j] = 1
if bfs(src):
return 2
return 1
ob = Solution()
matrix = [
[0, 0, 0, 0],
[0, 1, 0, 0],
[0, 1, 1, 0],
[0, 0, 0, 0],
]
print(ob.solve(matrix))Input
[ [0, 0, 0, 0], [0, 1, 0, 0], [0, 1, 1, 0], [0, 0, 0, 0], ]
Output
2
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