Given an m x n grid of characters board and a string word, return trueifwordexists in the grid.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true
Example 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true
Example 3:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false
Pseudocode
- dfs recursion to recurse over matrix
- seen array to prevent infinite loop
- is bfs possible?
Solution
varexist=function (board, word) {constseen= [];for (let i =0; i <board.length; i++) {seen.push(newArray(board[0].length).fill(false)); }functionwalk(x, y, idx) {if (x <0|| y <0|| x >=board.length|| y >= board[0].length) {// console.log('out of bounds')returnfalse; }// console.log(x, y, idx, word[idx])if (seen[x][y]) {// console.log('seen')return; }if (board[x][y] === word[idx]) {// console.log('found letter, incrementing') idx++; } else {// console.log('did not find letter')return; }if (idx ===word.length) {returntrue; }// pre seen[x][y] =true;// recurseconstdown=walk(x +1, y, idx);constright=walk(x, y +1, idx);constup=walk(x -1, y, idx);constleft=walk(x, y -1, idx);//postif (down || right || up || left) {// console.log('found path')returntrue; }// console.log('did not find path') seen[x][y] =false;returnfalse; }// return walk(0, 0, 0)for (let m =0; m <board.length; m++) {for (let n =0; n < board[0].length; n++) {if (walk(m, n,0)) {returntrue; } } }returnfalse;};