# Shortest distance between two cells in a matrix or grid

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; ` `} ` |

*chevron_right*

*filter_none*

Output:

6

This article is contributed by **Prashant Singh**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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