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.
s represents ‘source’
d represents ‘destination’
* represents cell you can travel
0 represents cell you can not travel
This problem is meant for single source and destination.
Examples:
Input : {'0', '*', '0', 's'},
{'*', '0', '*', '*'},
{'0', '*', '*', '*'},
{'d', '*', '*', '*'}
Output : 6
Input : {'0', '*', '0', 's'},
{'*', '0', '*', '*'},
{'0', '*', '*', '*'},
{'d', '0', '0', '0'}
Output : -1
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.
- 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:
6
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