Given a matrix of N*M order. Find the shortest distance from a source cell to a destination cell, traversing through limited cells only. Also you can move only up, down, left and right. If found output the distance else -1.Â
s represents 'source'Â
d represents 'destination'Â
* represents cell you can travelÂ
0 represents cell you can not travelÂ
This problem is meant for single source and destination.
Examples:Â
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Input : {'0', '*', '0', 's'},
{'*', '0', '*', '*'},
{'0', '*', '*', '*'},
{'d', '*', '*', '*'}
Output : 6
Input : {'0', '*', '0', 's'},
{'*', '0', '*', '*'},
{'0', '*', '*', '*'},
{'d', '0', '0', '0'}
Output : -1
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The idea is to BFS (breadth first search) on matrix cells. Note that we can always use BFS to find shortest path if graph is unweighted.Â
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- Store each cell as a node with their row, column values and distance from source cell.
- Start BFS with source cell.
- Make a visited array with all having "false" values except '0'cells which are assigned "true" values as they can not be traversed.
- Keep updating distance from source value in each move.
- Return distance when destination is met, else return -1 (no path exists in between source and destination).
// C++ Code implementation for above problem
#include <bits/stdc++.h>
using namespace std;
#define N 4
#define M 4
// QItem for current location and distance
// from source location
class QItem {
public:
int row;
int col;
int dist;
QItem(int x, int y, int w)
: row(x), col(y), dist(w)
{
}
};
int minDistance(char grid[N][M])
{
QItem source(0, 0, 0);
// To keep track of visited QItems. Marking
// blocked cells as visited.
bool visited[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
{
if (grid[i][j] == '0')
visited[i][j] = true;
else
visited[i][j] = false;
// Finding source
if (grid[i][j] == 's')
{
source.row = i;
source.col = j;
}
}
}
// applying BFS on matrix cells starting from source
queue<QItem> q;
q.push(source);
visited[source.row][source.col] = true;
while (!q.empty()) {
QItem p = q.front();
q.pop();
// Destination found;
if (grid[p.row][p.col] == 'd')
return p.dist;
// moving up
if (p.row - 1 >= 0 &&
visited[p.row - 1][p.col] == false) {
q.push(QItem(p.row - 1, p.col, p.dist + 1));
visited[p.row - 1][p.col] = true;
}
// moving down
if (p.row + 1 < N &&
visited[p.row + 1][p.col] == false) {
q.push(QItem(p.row + 1, p.col, p.dist + 1));
visited[p.row + 1][p.col] = true;
}
// moving left
if (p.col - 1 >= 0 &&
visited[p.row][p.col - 1] == false) {
q.push(QItem(p.row, p.col - 1, p.dist + 1));
visited[p.row][p.col - 1] = true;
}
// moving right
if (p.col + 1 < M &&
visited[p.row][p.col + 1] == false) {
q.push(QItem(p.row, p.col + 1, p.dist + 1));
visited[p.row][p.col + 1] = true;
}
}
return -1;
}
// Driver code
int main()
{
char grid[N][M] = { { '0', '*', '0', 's' },
{ '*', '0', '*', '*' },
{ '0', '*', '*', '*' },
{ 'd', '*', '*', '*' } };
cout << minDistance(grid);
return 0;
}
Output:Â
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6Please refer complete article on Shortest distance between two cells in a matrix or grid for more details!