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).
CPP
// 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 locationclass 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 codeint main(){ char grid[N][M] = { { '0', '*', '0', 's' }, { '*', '0', '*', '*' }, { '0', '*', '*', '*' }, { 'd', '*', '*', '*' } }; cout << minDistance(grid); return 0;} |
Java
/*package whatever //do not write package name here */// Java Code implementation for above problemimport java.util.LinkedList;import java.util.Queue;import java.util.Scanner;// QItem for current location and distance// from source locationclass QItem { int row; int col; int dist; public QItem(int row, int col, int dist) { this.row = row; this.col = col; this.dist = dist; }}public class Maze { private static int minDistance(char[][] grid) { QItem source = new QItem(0, 0, 0); // To keep track of visited QItems. Marking // blocked cells as visited. firstLoop: for (int i = 0; i < grid.length; i++) { for (int j = 0; j < grid[i].length; j++) { // Finding source if (grid[i][j] == 's') { source.row = i; source.col = j; break firstLoop; } } } // applying BFS on matrix cells starting from source Queue<QItem> queue = new LinkedList<>(); queue.add(new QItem(source.row, source.col, 0)); boolean[][] visited = new boolean[grid.length][grid[0].length]; visited[source.row][source.col] = true; while (queue.isEmpty() == false) { QItem p = queue.remove(); // Destination found; if (grid[p.row][p.col] == 'd') return p.dist; // moving up if (isValid(p.row - 1, p.col, grid, visited)) { queue.add(new QItem(p.row - 1, p.col, p.dist + 1)); visited[p.row - 1][p.col] = true; } // moving down if (isValid(p.row + 1, p.col, grid, visited)) { queue.add(new QItem(p.row + 1, p.col, p.dist + 1)); visited[p.row + 1][p.col] = true; } // moving left if (isValid(p.row, p.col - 1, grid, visited)) { queue.add(new QItem(p.row, p.col - 1, p.dist + 1)); visited[p.row][p.col - 1] = true; } // moving right if (isValid(p.row - 1, p.col + 1, grid, visited)) { queue.add(new QItem(p.row, p.col + 1, p.dist + 1)); visited[p.row][p.col + 1] = true; } } return -1; } // checking where it's valid or not private static boolean isValid(int x, int y, char[][] grid, boolean[][] visited) { if (x >= 0 && y >= 0 && x < grid.length && y < grid[0].length && grid[x][y] != '0' && visited[x][y] == false) { return true; } return false; } // Driver code public static void main(String[] args) { char[][] grid = { { '0', '*', '0', 's' }, { '*', '0', '*', '*' }, { '0', '*', '*', '*' }, { 'd', '*', '*', '*' } }; System.out.println(minDistance(grid)); }}// This code is contributed by abhikelge21. |
Output:
6
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