function activityNotifications(expenditure, d) {
  // Write your code here
  let notificationCounter = 0;
  const middle = Math.ceil(d / 2);
  
  // sort first trailing days, use this as base for future operations and to calculate the median
  const sortedTrailingDays = expenditure
    .slice(0, d)
    .sort((a, b) => (b < a ? 0 : -1));

  for (let i = d; i < expenditure.length; i++) {
    const spentThisDay = expenditure[i];

    if (i > d) { // ignore first iteration, trailings days are sorted once before the loop
      /**
       * 2 operations on the sorted array of trailing day expenditures
       *    1. remove the expenditure of the day that dropped out of the trailing days
       *    2. add the expenditure of the previous day to the trailing days at the correct position
       * 
       * this is way faster than sorting the whole array on every iteration 
       * with the js-array-sort method (as done above the for-loop)
       *  -> this way the function was fast enough to succeed all test-cases, 
       *     even in slow-ass single-CPU-javascript :D
       */
      // remove "dropped out" number/day
      sortedTrailingDays.splice(
        sortedTrailingDays.indexOf(expenditure[i - d - 1]),
        1
      );
      // insert "previous-day-expenditure" to sorted array at the correct position
      const spentDayBefore = expenditure[i - 1];
      if (spentDayBefore <= sortedTrailingDays[0]) {
        sortedTrailingDays.unshift(spentDayBefore); // first pos
      } else if (
        spentDayBefore >= sortedTrailingDays[sortedTrailingDays.length - 1]
      ) {
        sortedTrailingDays.push(spentDayBefore);  // last pos
      } else {
        // find first higher number -> insert there
        const idx = sortedTrailingDays.indexOf(
          sortedTrailingDays.find((n) => n > spentDayBefore) // most likely to slow for testcase-5
        );
        sortedTrailingDays.splice(idx, 0, spentDayBefore);
      }
    }
    
    // calculate median
    const median = d % 2 === 0 ? 
        (sortedTrailingDays[middle] + sortedTrailingDays[middle - 1]) / 2 
        : sortedTrailingDays[middle - 1];

    if (spentThisDay >= median * 2) {
      notificationCounter++;
    }
  }

  return notificationCounter;
}