Question 189301 in Javascript Question posted in
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Javascript – Check if all object in array is greater than x number

I have an array of objects representing sales data for different cities. Each city has an array of rooms with corresponding counts. I want to check if all cities have a count of at least three for rooms 2, 3, and 4. If this condition is met, I would like to obtain the names of all the cities that satisfy it.

Here is an example of the data structure:

const sales = {
  "City 1": [
    {
      "rooms": 1,
      "count": 1
    },
    {
      "rooms": 2,
      "count": 2
    },
    {
      "rooms": 3,
      "count": 3
    }
  ],
  "City 2": [
    {
      "rooms": 1,
      "count": 1
    },
    {
      "rooms": 2,
      "count": 1
    },
    {
      "rooms": 3,
      "count": 1
    },
    {
      "rooms": 4,
      "count": 2
    }
  ],
  "City 3": [
    {
      "rooms": 2,
      "count": 6
    },
    {
      "rooms": 4,
      "count": 7
    }
  ],
  "City 4": [
    {
      "rooms": 1,
      "count": 4
    },
    {
      "rooms": 2,
      "count": 6
    },
    {
      "rooms": 3,
      "count": 3
    },
    {
      "rooms": 4,
      "count": 7
    }
  ]
};

Several solutions I tried

for (const city in sales) {
    let isPass = true;
    for (const room of sales[city]) {
      if (room.count < x) {
        isPass = false;
        break;
      }
    }
    if (isPass) {
      // do stuff here
      return true;
    }
  }

Also tried

let filteredCities = [];

  for (const city in sales) {
    const roomsData = sales[city];

    const hasRoom2 = roomsData.some((room) => room.rooms === 2);
    const hasRoom3 = roomsData.some((room) => room.rooms === 3);
    const hasRoom4 = roomsData.some((room) => room.rooms === 4);

    const isCount = roomsData.every(
      (room) => room.count >= 3
    );

    if (hasRoom2 && hasRoom3 && hasRoom4 && isCount) {
      filteredCities.push(city);
    }
  }

  return filteredCities;

I need assistance in writing a JavaScript function that will perform this check and return the following:

A boolean value indicates whether all cities have at least three counts for rooms 2, 3, and 4.
If the boolean value is true, a list of the names of the cities that satisfy the condition.

Tags:

2

Answers


  1. 0 
    const sales = {
  "City 1": [{
      "rooms": 1,
      "count": 1
    },
    {
      "rooms": 2,
      "count": 2
    },
    {
      "rooms": 3,
      "count": 3
    }
  ],
  "City 2": [{
      "rooms": 1,
      "count": 1
    },
    {
      "rooms": 2,
      "count": 1
    },
    {
      "rooms": 3,
      "count": 1
    },
    {
      "rooms": 4,
      "count": 2
    }
  ],
  "City 3": [{
      "rooms": 1,
      "count": 4
    },
    {
      "rooms": 2,
      "count": 6
    },
    {
      "rooms": 3,
      "count": 3
    },
    {
      "rooms": 4,
      "count": 7
    }
  ]

}

const listOfCityWithAllCountGT3 = []

for (const key in sales) {
  if (sales[key].some(e => e.count < 3)) {
  //here array.some checks if any element is smaller than 3
    continue;
  } else {
    listOfCityWithAllCountGT3.push(key)
  }

}

console.log(listOfCityWithAllCountGT3)
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  2. 0
    1. I want to check each city independently and select the one that passes the test.
    2. Each count is >=3 in rooms 2, 3, and 4.

    So when you’re checking each city, you just need to keep track of records that meet the requirements. You can do this by using a Set filled with relevant rooms numbers. As you loop through the records, for each record, check if rooms is both present in the set and satisfies the condition. If true, delete it from the set. When the set becomes empty, you are sure that the current city passed the test. Refill the set for each subsequent city and repeat.

    const sales = {
  "City 1": [
    { "rooms": 1, "count": 1 },
    { "rooms": 2, "count": 2 },
    { "rooms": 3, "count": 3 }
  ],
  "City 2": [
    { "rooms": 1, "count": 1 },
    { "rooms": 2, "count": 1 },
    { "rooms": 3, "count": 1 },
    { "rooms": 4, "count": 2 }
  ],
  "City 3": [
    { "rooms": 2, "count": 6 },
    { "rooms": 4, "count": 7 }
  ],
  "City 4": [
    { "rooms": 1, "count": 4 },
    { "rooms": 2, "count": 6 },
    { "rooms": 3, "count": 3 },
    { "rooms": 4, "count": 7 }
  ]
};

function filterLocations (locations, rooms, minCount) {

    let names = Object.keys(locations);
    let found = [];

    for (let i = 0; i < names.length; i++) {

        let location = locations[names[i]];
        if (!Array.isArray(location)) {
            continue;
        }

        let set = new Set(rooms);
        for (let j = 0; j < location.length; j++) {
            let entry = location[j];
            if (set.has(entry?.rooms) && entry?.count >= minCount) {
                set.delete(entry?.rooms);
            }
            if (set.size == 0) {
                found.push(names[i]);
                break;
            }
        }
    }

    return found;
}

console.log(filterLocations(sales, [2, 3, 4], 3));
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