Word Search

Word Search - LeetCodeLeetCode

Problem

Given an m x n grid of characters board and a string word, return true if word exists 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

var exist = function (board, word) {
  const seen = [];

  for (let i = 0; i < board.length; i++) {
    seen.push(new Array(board[0].length).fill(false));
  }

  function walk(x, y, idx) {
    if (x < 0 || y < 0 || x >= board.length || y >= board[0].length) {
      // console.log('out of bounds')
      return false;
    }

    // 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) {
      return true;
    }

    // pre
    seen[x][y] = true;
    // recurse
    const down = walk(x + 1, y, idx);
    const right = walk(x, y + 1, idx);
    const up = walk(x - 1, y, idx);
    const left = walk(x, y - 1, idx);
    //post
    if (down || right || up || left) {
      // console.log('found path')
      return true;
    }

    // console.log('did not find path')
    seen[x][y] = false;
    return false;
  }
  // 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)) {
        return true;
      }
    }
  }

  return false;
};

Time and Space Complexity

Time

• What did the code do

• Total -

Space

• What did the code do

• Total -