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

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