JavaScript TicTacToe if… Winner detection

Question

I just have a small problem.

The final assignment in my computer science class is to write a game of Tic-Tac-Toe in JavaScript. The game works by clicking on cells w

Answer 1

Assume 'squares' is a Array of Arrays of square size with 0 for not filled , 1 for player 1 (x) and -1 for the other player (O).

e.g.

let squares= Array(3).fill(Array(3).fill(0))

would be the staring board.

The following works for board size 1,2,3,4... where for a board of size n, n consecutive x's or o's need to be in a row, column or diagonal.

Returns 0 if nobody won yet and 1 for player 1 and -1 for the other player.

First defining two helper functions to make the code more readable.

const add = (a, b) => a + b

function sum(array){  
   return array.reduce(add);
}

function calculateWinner(squares) {
  // check for horizontal wins along rows and diagonals
  let winner = calculateWinnerInner(squares);
  if (winner !== 0) return winner;
  // check for possible vertical wins as well
  const stranspose = squares.map((col, i) => squares.map(row => row[i]));
  return calculateWinnerInner(stranspose);
}
 
function calculateWinnerInner(squares) {
  for (let r = 0; r < squares.length; r++) {
    if (squares[r].length === sum(squares[r])) {
      return 1;
    }
    if (squares[r].length === - sum(squares[r])) {
      return -1;
    }
  }

  const diagonal = squares.map((row, r) => squares[r][r]);

  if (squares[0].length === sum(diagonal)) {
    return 1;
  }
  if (squares[0].length === -sum(diagonal)) {
    return -1;
  }
  
 const len=squares.length;
 const crossdiagonal = squares.map((row, r) => squares[r][len-r-1]);

 if (squares[0].length === sum(crossdiagonal)) {
  return 1;
 }
 if (squares[0].length === -sum(crossdiagonal)) {
  return -1;
 }

  return 0;
}