interview questions – Square Spiral Matrix in Javascript

I’ve been grinding some Leetcode for an interview coming up and just completed Square Spiral in Javascript.

Looking for feedback on performance. This ranked faster than 58% of submissions, would there be anyway to increase the speed using a similar implementation to what I have here?

enter image description here

let matrix = (
  (1, 2, 3, 4,  5),
  (6, 7, 8, 9, 10),
  (11,12,13,14,15),
  (16,17,18,19,20),
  (21,22,23,24,25)
)

var spiralOrder = (matrix) => {
  //make copy
  var dup=JSON.parse(JSON.stringify(matrix));
  //setup variables
  let maxX, maxY, baseX, baseY, x, y, vecX, vecY, nums, totalSquares;

  //Max indexes for array and sub arrays
  maxY=matrix(0).length-1;
  maxX=matrix.length-1;

  //baseX tells x when it has reached the first sub array.
  baseX=0;
  
  //baseY tells y when it has reached the first index of a sub array.
  baseY=0;
    
  //our indexes
  x=0;
  y=0;
    
  //our movement vectors
  vecX=0;
  vecY=0;
    
  //the count for keeping track of positions iterated
  count=0;
    
  //the total amount of squares to be iterated
  totalSquares=matrix.length*matrix(0).length;
    
  //return value array
  nums=();

  //I don't get how subArrays with only a single element could
  //still be considered spiralable, but this if-statement handles
  //those edge cases. Thanks Leetcode.
  if (matrix(0).length===1) {
    for (var i=0;i<matrix.length;i++) {
      nums.push(matrix(i)(0))
    }
    return nums
  }
    
  //While our iterated count is not equal to the total amount of
  //squares to be iterated. When this is true we know we
  //have completed the spiral.
  while(count!==totalSquares) {
      
    nums.push(dup(x)(y));
    count++;

    //Our loop iterates around in a spiral. On every turn
    //the loop increases the opposite side baseX, baseY,
    //maxX or maxY to increase the tightness of the spiral
    //as it loops.
      
    //This handles starting the very first iteration, moving right.
    if (x===0&&y===0) {
      vecX=0;
      vecY=1;
    //We've reached the top rightmost square, move down.
    } else if (x===baseX&&y===maxY) {
      vecX=1;
      vecY=0;

      //This is a little confusing. We can't increment baseY
      //until after the first iteration through the
      //first sub array is complete. This if statement
      //confirms that we're clear to increment baseY.
      if (baseX>0) {
        baseY+=1;
      }
    //We've reached bottom rightmost square, move left, increment baseX. 
    } else if (x===maxX&&y===maxY) {
      vecX=0;
      vecY=-1;
      baseX+=1;
    //We've reached bottom leftmost square, move up, decrement maxY.
    } else if (x===maxX&&y===baseY) {
      vecX=-1;
      vecY=0
      maxY-=1;
    //We've reached top leftmost square, move right, decrement maxX.
    } else if (x===baseX&&y===baseY) {
      vecX=0;
      vecY=1;
      maxX-=1;
    }

    //Increment x or y by the vector.
    x+=vecX;
    y+=vecY;
  }

  return nums
}

```