Cat and Mouse LeetCode solution in JavaScript

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var catMouseGame = function (graph) {
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  let n = graph.length,
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    memo = [...Array(n + 1)].map((d) =>
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      [...Array(n + 1)].map((d) => [...Array(2 * n + 1)])
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    ),
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    seen = [...Array(n + 1)].map((d) =>
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      [...Array(n + 1)].map((d) => [...Array(2)])
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    );
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  //dfs returns 0 1 2, whether the current player loses,wins, or draws respectively
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  let dfs = (M, C, level) => {
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    let turn = level % 2,
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      curr = turn ? C : M,
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      draw = 0,
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      res = 0;
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    //draw when we ve seen the state before or cycles
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    if (seen[M][C][turn] !== undefined || level >= 2 * n)
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      return (memo[M][C][level] = 2);
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    if (M == 0)
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      // win for mouse if it reaches the hole, loss for cat
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      memo[M][C][level] = turn ^ 1;
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    if (M == C)
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      // win for cat if it reaches the mouse, loss for mouse
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      memo[M][C][level] = turn;
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    if (memo[M][C][level] === undefined) {
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      seen[M][C][turn] = 0; //set this state as visited
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      for (
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        let i = 0;
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        i < graph[curr].length && !res;
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        i++ //traverse for the available edges
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      )
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        if (!(turn && !graph[curr][i])) {
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          //The cat cant move into the hole
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          let val = turn
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            ? dfs(M, graph[curr][i], level + 1)
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            : dfs(graph[curr][i], C, level + 1);
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          if (val === 2)
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            draw = 1; //set draw as an available option
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          else res |= 1 ^ val; //minimax logic, always prefer the losing state of the opponent
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        }
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      memo[M][C][level] = res || 2 * draw; // set in this order 1->2->0
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    }
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    seen[M][C][turn] = undefined; // de-set the state for the current game,as it concluded
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    return memo[M][C][level];
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  };
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  return [2, 1, 0][dfs(1, 2, 0)]; //js eye candy
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};