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Find the minimum number of moves needed to move from one cell of matrix to another in Python
Suppose we have one N X N matrix M, and this is filled with 1, 0, 2, 3, We have to find the minimum numbers of moves required to move from source cell to destination cell. While visiting through blank cells only, we can visit up, down, right and left.
Cell with value 1 indicates Source.
Cell with value 2 indicates Destination.
Cell with value 3 indicates Blank cell.
Cell with value 0 indicates Blank Wall.
There will be only one source and only one destination cells. There may be more than one path to reach destination from source cell. Now, each move in matrix we consider as '1'.
So, if the input is like
| 3 | 3 | 1 | 0 |
| 3 | 0 | 3 | 3 |
| 3 | 3 | 0 | 3 |
| 0 | 3 | 2 | 3 |
then the output will be 5,
| 3 | 3 | 1 | 0 |
| 3 | 0 | 3 | 3 |
| 3 | 3 | 0 | 3 |
| 0 | 3 | 2 | 3 |
From start to destination the green path is shortest.
To solve this, we will follow these steps −
nodes := order * order + 2
g := a blank graph with ‘nodes’ number of vertices
k := 1
for i in range 0 to order, do
for j in range 0 to order, do
if mat[i, j] is not same as 0, then
if is_ok (i , j + 1 , mat) is non-zero, then
create an edge between k and k + 1 nodes of g
if is_ok (i , j - 1 , mat) is non-zero, then
create an edge between k, k - 1 nodes of g
if j < order - 1 and is_ok (i + 1 , j , mat) is non-zero, then
create an edge between k, k + order nodes of g
if i > 0 and is_ok (i - 1 , j , mat) is non-zero, then
create an edge between k, k - order nodes of g
if mat[i, j] is same as 1, then
src := k
if mat[i, j] is same as 2, then
dest := k
k := k + 1
return perform bfs from src to dest of g
Example
Let us see the following implementation to get better understanding −
class Graph: def __init__(self, nodes): self.nodes = nodes self.adj = [[] for i in range(nodes)] def insert_edge (self, src , dest): self.adj[src].append(dest) self.adj[dest].append(src) def BFS(self, src, dest): if (src == dest): return 0 level = [-1] * self.nodes queue = [] level[src] = 0 queue.append(src) while (len(queue) != 0): src = queue.pop() i = 0 while i < len(self.adj[src]): if (level[self.adj[src][i]] < 0 or level[self.adj[src][i]] > level[src] + 1 ): level[self.adj[src][i]] = level[src] + 1 queue.append(self.adj[src][i]) i += 1 return level[dest] def is_ok(i, j, mat): global order if ((i < 0 or i >= order) or (j < 0 or j >= order ) or mat[i][j] == 0): return False return True def get_min_math(mat): global order src , dest = None, None nodes = order * order + 2 g = Graph(nodes) k = 1 for i in range(order): for j in range(order): if (mat[i][j] != 0): if (is_ok (i , j + 1 , mat)): g.insert_edge (k , k + 1) if (is_ok (i , j - 1 , mat)): g.insert_edge (k , k - 1) if (j < order - 1 and is_ok (i + 1 , j , mat)): g.insert_edge (k , k + order) if (i > 0 and is_ok (i - 1 , j , mat)): g.insert_edge (k , k - order) if(mat[i][j] == 1): src = k if (mat[i][j] == 2): dest = k k += 1 return g.BFS (src, dest) order = 4 mat = [[3,3,1,0], [3,0,3,3], [3,3,0,3], [0,3,2,3]] print(get_min_math(mat))
Input
[[3,3,1,0], [3,0,3,3], [3,3,0,3], [0,3,2,3]]
Output
0
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