Rendezvous with Cassidoo #186 - Interview question of the week - validTTTPosition

validTTTPosition.js

const validTTTPosition = (boardInput) => {
  // Assuming boardInput of form ["XOX", " X ", "   "]
  // Re-organise board chars; I'm comfy with arrays.
  const boardChars = [...boardInput.join("")];
  // Player char counts
  const Xcount = boardChars.filter((char) => char === "X").length;
  const Ocount = boardChars.filter((char) => char === "O").length;
  // Mustn't have less Xs than Os; assume X goes first
  if (Xcount < Ocount) return false;
  // Mustn't have 2 more Xs than Os; assume we take turns
  if (Xcount - Ocount > 1) return false;
  // Ok, let's just compute the 8 board "words" up front #noPerf 🤷
  const boardWords = [
    `${boardChars[0]}${boardChars[1]}${boardChars[2]}`, // Horizontal
    `${boardChars[3]}${boardChars[4]}${boardChars[5]}`, // Horizontal
    `${boardChars[6]}${boardChars[7]}${boardChars[8]}`, // Horizontal
    `${boardChars[0]}${boardChars[3]}${boardChars[6]}`, // Vertical
    `${boardChars[1]}${boardChars[4]}${boardChars[7]}`, // Vertical
    `${boardChars[2]}${boardChars[5]}${boardChars[8]}`, // Vertical
    `${boardChars[0]}${boardChars[4]}${boardChars[8]}`, // Diagonal
    `${boardChars[2]}${boardChars[4]}${boardChars[6]}`, // Diagonal
  ];
  // If Xs win there must be one more Xs, than Os, on the board.
  if (boardWords.includes("XXX") && Xcount !== Ocount + 1) {
    return false;
  }
  // If Os win there must be equal counts of Xs and Os on the board.
  if (boardWords.includes("OOO") && Xcount !== Ocount) {
    return false;
  }
  // Exhausted checks, assume board position is valid 😅
  return true;
};