PROBLEM DESCRIPTION
Given a grid of size n*n filled with 0, 1, 2, 3. Check whether there is a path possible from the source to destination. You can traverse up, down, right and left. The description of cells is as follows:
- A value of cell 1 means Source.
- A value of cell 2 means Destination.
- A value of cell 3 means Blank cell.
- A value of cell 0 means Wall (blocked cell which we cannot traverse).
Note: There are only a single source and a single destination.
SOLUTION
This can be solved using DFS.
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class Solution {
public boolean is_Possible(int[][] grid) {
int n = grid.length;
int m = grid[0].length;
// Initialize source and destination coordinates
int si = -1, sj = -1, di = -1, dj = -1;
// get the source and destination
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) {
si = i;
sj = j;
} else if (grid[i][j] == 2) {
di = i;
dj = j;
}
// Break if both source and destination are found
if (si != -1 && di != -1) {
break;
}
}
}
// Perform depth-first search (DFS) to check if a path exists
return dfs(grid, si, sj, di, dj);
}
public boolean dfs(int[][] grid, int si, int sj, int di, int dj) {
// Check if the current cell is out of bounds or a wall
if (di < 0 || dj < 0 || di >= grid.length || dj >= grid[0].length || grid[di][dj] == 0) {
return false;
}
// If the current cell is the source cell, path is found
if (di == si && dj == sj) {
return true;
}
// Otherwise, mark the current cell as visited by setting it to 0
grid[di][dj] = 0;
// possible moves (right, left, down, up)
int[] x = {0, 0, 1, -1};
int[] y = {1, -1, 0, 0};
// init:
boolean reached = false;
// Explore all four possible directions
for (int k = 0; k < 4; k++) {
int nextX = di + x[k];
int nextY = dj + y[k];
// Recursively perform DFS on the next cell
// reached will be set to true if any of the moves returns true
reached |= dfs(grid, si, sj, nextX, nextY);
}
return reached;
}
}