To successfully achieve my goal in JavaScript, I must generate an array containing weekdays only when there are multiple identical values present

Question

To successfully achieve my goal in JavaScript, I must generate an array containing weekdays only when there are multiple identical values present

Explaining this concept might be a bit complex, but here is the official breakdown:

mostPopularDays: This function identifies which day of the week had the highest number of visitors at the pet store. If multiple days tie for the highest traffic, an array containing those days should be returned in any order. In case of null or an empty input array, the function should return null. The input consists of Weekday objects created using the prototype function outlined in petstore.js. The desired output is a string with the name of the most popular day if only one stands out, and an array of names (in string format) for multiple top days.

The current version of the code snippet looks like this:

function mostPopularDays (week) {
  var name,
      dayInstance,
      highestTrafficYet = -1;

  for (var i = 0; i < week.length; ++i) {
    dayInstance = week[i];
    traffic = dayInstance.traffic;

    if (highestTrafficYet < traffic) {
      name = dayInstance.name;
      highestTrafficYet = traffic;
    }
  }
  return name;
}

If there are multiple equally popular days, how can I modify the code to return an array instead?

javascript arrays sorting

Answer 1

Answer №1

Refer to the comments below for a detailed explanation. By sorting the values initially, you can easily identify the day with the highest traffic. Utilize this high-traffic number to filter out any days with lower traffic. Subsequently, you can either output a single name if there is only one busiest day, or map the results to display only the names if there are multiple.

let data = [{
  traffic: 4, name: 'mon'
}, {
  traffic: 2, name: 'weds'
}, {
  traffic: 4, name: 'tues'
}];

function mostPopularDays(week) {
  let sorted = week.sort((a, b) => b.traffic - a.traffic); // organize the top traffic days
  let highest = sorted[0].traffic // take the highest value as a sample
  days = sorted.filter(a => a.traffic >= highest); // keep only the high traffic days in the list
  if (days.length === 1) return days[0].name // return the name if there is only one
  else return days.map(e => e.name) // if multiple, create an array of day names
}
console.log(mostPopularDays(data))

Answer 2

Refer to the comments below for a detailed explanation. By sorting the values initially, you can easily identify the day with the highest traffic. Utilize this high-traffic number to filter out any days with lower traffic. Subsequently, you can either output a single name if there is only one busiest day, or map the results to display only the names if there are multiple.

let data = [{
  traffic: 4, name: 'mon'
}, {
  traffic: 2, name: 'weds'
}, {
  traffic: 4, name: 'tues'
}];

function mostPopularDays(week) {
  let sorted = week.sort((a, b) => b.traffic - a.traffic); // organize the top traffic days
  let highest = sorted[0].traffic // take the highest value as a sample
  days = sorted.filter(a => a.traffic >= highest); // keep only the high traffic days in the list
  if (days.length === 1) return days[0].name // return the name if there is only one
  else return days.map(e => e.name) // if multiple, create an array of day names
}
console.log(mostPopularDays(data))

Answer 3

Answer №2

Start by identifying the highest number of visitors on any given day, then utilize Array#filter to locate all days with that level of traffic.

function findBusiestDays (week) {
  if(!week?.length) return null;
  const max = Math.max(...week.map(x => x.traffic));
  const busyDays = week.filter(x => x.traffic === max).map(x => x.name);
  return busyDays.length === 1 ? busyDays[0]: busyDays;
}

Answer 4

Start by identifying the highest number of visitors on any given day, then utilize Array#filter to locate all days with that level of traffic.

function findBusiestDays (week) {
  if(!week?.length) return null;
  const max = Math.max(...week.map(x => x.traffic));
  const busyDays = week.filter(x => x.traffic === max).map(x => x.name);
  return busyDays.length === 1 ? busyDays[0]: busyDays;
}

Answer 5

Answer №3

This function is designed to determine the most popular days based on traffic:

function mostPopularDays(week) {
  var name,
    dayInstance,
    highestTrafficYet = -1,
    trafficIndexedDaysObj = {};

  for (var i = 0; i < week.length; ++i) {
    dayInstance = week[i];
    traffic = dayInstance.traffic;

    if (highestTrafficYet < traffic) {
      name = dayInstance.name;
      highestTrafficYet = traffic;

      // Clear previous hightest days
      trafficIndexedDaysObj = {};

      // Store highest Day in Array and put the Array in an Object with Traffic as key
      trafficIndexedDaysObj[traffic] = [name];
    } else if (highestTrafficYet == traffic && Array.isArray(trafficIndexedDaysObj[traffic])) {
      // If the same Traffic exists, append that Day to the already created Days Array
      trafficIndexedDaysObj[traffic].push(dayInstance.name);
    }
  }

  if (Object.keys(trafficIndexedDaysObj).length > 0) {
    // Return array of Days when the Highest Traffic occurred on more than one Days 
    var keys = Object.keys(trafficIndexedDaysObj);

    if (trafficIndexedDaysObj[keys[0]].length > 1) {
      return trafficIndexedDaysObj[keys[0]];
    }
  }

  return name;
}

mostPopularDays([{
  'traffic': 100,
  'name': 'Monday'
}, {
  'traffic': 64,
  'name': 'Tuesday'
}, {
  'traffic': 52,
  'name': 'Wednesday'
}, {
  'traffic': 100,
  'name': 'Thursday'
}, {
  'traffic': 100,
  'name': 'Friday'
}, {
  'traffic': 100,
  'name': 'Saturday'
}])

View the full code implementation here

Answer 6

This function is designed to determine the most popular days based on traffic:

function mostPopularDays(week) {
  var name,
    dayInstance,
    highestTrafficYet = -1,
    trafficIndexedDaysObj = {};

  for (var i = 0; i < week.length; ++i) {
    dayInstance = week[i];
    traffic = dayInstance.traffic;

    if (highestTrafficYet < traffic) {
      name = dayInstance.name;
      highestTrafficYet = traffic;

      // Clear previous hightest days
      trafficIndexedDaysObj = {};

      // Store highest Day in Array and put the Array in an Object with Traffic as key
      trafficIndexedDaysObj[traffic] = [name];
    } else if (highestTrafficYet == traffic && Array.isArray(trafficIndexedDaysObj[traffic])) {
      // If the same Traffic exists, append that Day to the already created Days Array
      trafficIndexedDaysObj[traffic].push(dayInstance.name);
    }
  }

  if (Object.keys(trafficIndexedDaysObj).length > 0) {
    // Return array of Days when the Highest Traffic occurred on more than one Days 
    var keys = Object.keys(trafficIndexedDaysObj);

    if (trafficIndexedDaysObj[keys[0]].length > 1) {
      return trafficIndexedDaysObj[keys[0]];
    }
  }

  return name;
}

mostPopularDays([{
  'traffic': 100,
  'name': 'Monday'
}, {
  'traffic': 64,
  'name': 'Tuesday'
}, {
  'traffic': 52,
  'name': 'Wednesday'
}, {
  'traffic': 100,
  'name': 'Thursday'
}, {
  'traffic': 100,
  'name': 'Friday'
}, {
  'traffic': 100,
  'name': 'Saturday'
}])

View the full code implementation here

To successfully achieve my goal in JavaScript, I must generate an array containing weekdays only when there are multiple identical values present

Answer №1

Answer №2

Answer №3

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