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.

> 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.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

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));

<!-- end snippet -->

I find it cleaner to separate out the code that finds the rooms by id from the main code. I think it makes for more readable blocks.

Also, it wasn't clear to me what output you wanted. So my version splits out finding the object entries that match your criteria from the formatting of their results. If you wanted the output to look like the input object, just filtered to keep those that match your criteria, I would use this:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

const findRoom = (city, roomNbr) =>
city.find(({rooms}) =>  rooms == roomNbr)
const findMatchingCities = (cities) => Object.entries(cities).filter (
([name, city]) => [2, 3, 4].every(roomNbr => findRoom(city, roomNbr)?.count >= 3)
)
const selectCities = (cities) => 
Object.fromEntries(findMatchingCities(cities))
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}]}
console .log (selectCities(sales))

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But if you just wanted the matching ciy names, I would use this function instead:

const selectCityNames = (cities) => 
  findMatchingCities(cities).map(([name]) => name)

You can see it by expanding the following snippet:

<!-- begin snippet: js hide: true console: true babel: false -->

<!-- language: lang-js -->

const findRoom = (city, roomNbr) =>
city.find(({rooms}) =>  rooms == roomNbr)
const findMatchingCities = (cities) => Object.entries(cities).filter (
([name, city]) => [2, 3, 4].every(roomNbr => findRoom(city, roomNbr)?.count >= 3)
)
const selectCityNames = (cities) => 
findMatchingCities(cities).map(([name]) => name)
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}]}
console .log (selectCityNames(sales))

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.as-console-wrapper {max-height: 100% !important; top: 0}

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In either case, the main function, findMatchingCities filters the entries of our input object, using our findRoom for each of the room numbers we want to test (2, 3, 4), and checking if they have a count property of at least 2. It returns an array of object entries, each in the form ['CityName', [{rooms: 1, count: 1}, ...]] You could easily fold this into selectCities or selectCityNames and remove one function. Or not. This breakdown is fairly clean.

If the room numbers and/or the minimum count should be dynamic, it's trivial to add parameters to findMatchingCities and the function that calls it.

I found a solution,

I was able to solve it by iterating over each city and checking if the count for rooms 2, 3, and 4 is >= 3 in each city

Here:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

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 checkCity(cities) {
const availableCities = [];
for (const city in cities) {
const rooms = cities[city];
const room2 = rooms.find(room => room.rooms === 2);
const room3 = rooms.find(room => room.rooms === 3);
const room4 = rooms.find(room => room.rooms === 4);
if (room2 && room3 && room4 && room2.count >= 3 && room3.count >= 3 && room4.count >= 3) {
availableCities.push(city);
}
}
return availableCities;
}
console.log(checkCity(sales)); 

<!-- end snippet -->