Maximum Number of Visible Points LeetCode solution in JavaScript

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// basically First i transform my points into degrees in regards to my location as the center
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// I need to perform a sliding window to see how many points can fit within my window of length
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// angle
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var visiblePoints = function (points, angle, location) {
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  //Math.atan2(y,x) returns the ANGLE in RADIANS between the point (X,Y), the CENTER (0,0) and x'x
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  // so Math.atan2(5,5) * (180/Math.pi) === 45 //transforms it from radians to degrees instead
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  let [sx, sy] = location;
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  let onCenter = points.filter(([x, y]) => x == sx && y == sy).length; //if i m standing on a point i always count it
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  points = points
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    .filter(([x, y]) => !(x == sx && y == sy)) //other than that i dont want it messing with my result
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    .map(([x, y]) => {
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      // i need to transform my center from (0,0) to location, hence (y-sy,x-sx)
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      return Math.atan2(y - sy, x - sx) * (180 / Math.PI); //returns the degrees
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    })
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    .sort((a, b) => a - b);
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  //I will now perform the circular array duplication trick in order to consider points from different view
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  points = [...points, ...points.map((d) => d + 360)];
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  // so for example if a point is 340, it can go with a point that is 15
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  // example : [0,60,230,250,359], angle=200
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  // would become     [0,60,230,250,359,360,420,590,610,719]
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  // would allow me to pick  *   *   *   *   * ,which is practically [230,250,359,0,60]
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  // (every element of my starting array), which are obviously visible with anangle fo 200 deg
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  //now i will perform a sliding window that tracks the points visible from my current degree-my angle degrees
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  let start = 0,
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    n = points.length,
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    result = 0;
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  for (let end = 0; end < n; end++) {
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    while (start < end && points[start] < points[end] - angle)
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      //if the point of start is no bueno
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      start++; // shrink the window until all the points are within angle degrees
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    result = Math.max(result, end - start + 1);
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  }
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  return result + onCenter; //the points i can see + the ones i ms tanding on
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};