```
public class Solution {
private int width;
private int height;
private char[][] board;
private String word;
public boolean exist(char[][] brd, String wrd) {
// Store parameters that won't change
if((height = brd.length) == 0 ) return false;
if((width = brd[0].length)==0) return false;
if(wrd.length()==0) return true;
board = brd; word = wrd;
// Locate the root by looking
// for the 1st character
// in the board
for (int y = 0; y < height; y ++) {
for (int x = 0; x < width; x ++) {
if (board[y][x] == word.charAt(0))
if (dfs(y,x,1,new boolean[height][width]) == true) return true;
}
}
return false;
}
// Use DFS other than BFS, because the "visited" map is commonly used
public boolean dfs(int cy, int cx, int idx, boolean[][] visited) {
// If no more characters, the search is completed successfully
if (idx == word.length()) return true;
boolean found = false;
visited[cy][cx] = true;
// The part of code I do not feel like writing evertime ---------------------------->
// Check left neighbor if current not on the left boundary
// If word[idx] is found, proceed by a recursive call
if (cx > 0) {
if (visited[cy][cx-1] == false && board[cy][cx-1] == word.charAt(idx))
found = dfs(cy, cx-1, idx + 1, visited);
}
// Right neighbor
if (cx < width - 1 && found == false) {
if (visited[cy][cx+1] == false && board[cy][cx+1] == word.charAt(idx))
found = dfs(cy, cx+1, idx + 1, visited);
}
// Up neighbor
if (cy > 0 && found == false) {
if (visited[cy-1][cx] == false && board[cy-1][cx] == word.charAt(idx))
found = dfs(cy-1, cx, idx + 1, visited);
}
// Down neighbor
if (cy < height - 1&& found == false) {
if (visited[cy+1][cx] == false && board[cy+1][cx] == word.charAt(idx))
found = dfs(cy+1, cx, idx + 1, visited);
}
// <--------------------------------------------------------------------------------------------
// Undo visited to current tile if search wasn't successful
if (found == false) {
visited[cy][cx] = false;
} return found;
}
}
```

Common to questions related to graph searching and 2D arrays, there are x and y coordinates and 4 corners ULFR to search, so I cannot write code as cool as

```
for (Node n : (Iterable<Node>)neighbors) { //... }
```

to check all of them.

I tried mapping the 2D array to 1D as using

```
int id = x + board_width * y;
int x = id % board_width;
int y = id / board_width;
```

but that also makes the code complex and slightly added constant computation cost.

So is there any concise solution to code for graphs in 2D arrays?