Determine the minimum value of an object's keys by comparing them to a given number

Question

Determine the minimum value of an object's keys by comparing them to a given number

Consider the following scenario:

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const value = 15;

I am looking for an efficient way to compare the 'value' against the keys in the object and retrieve the corresponding value from the smaller range. In this case, the expected result would be b as 15 falls between 10 and 20.

This was my initial approach:

for(var i=0; i < keys.length; i++){
    const item = parseInt(keys[i],10);
    if (item == keys[i]) {
      return keys[i];
    }
  }

However, I realized that this method is not very efficient...

javascript arrays object ecmascript-6

Answer 1

Answer №1

To retrieve all the keys of an object and stop the loop when a certain condition is met, you can utilize Object.keys to extract the keys into a new array. Then, within the .some method (as recommended by @Keith), you can stop the loop by returning true once the current value exceeds a specified threshold set by the variable value. The final result will be stored in the variable named previous.

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const value = 15;
let previous = 0;

Object.keys(list).some(k => {
  if (k > value) {
    return true;
  } 
  previous = k
});

console.log(list[previous]);

Answer 2

To retrieve all the keys of an object and stop the loop when a certain condition is met, you can utilize Object.keys to extract the keys into a new array. Then, within the .some method (as recommended by @Keith), you can stop the loop by returning true once the current value exceeds a specified threshold set by the variable value. The final result will be stored in the variable named previous.

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const value = 15;
let previous = 0;

Object.keys(list).some(k => {
  if (k > value) {
    return true;
  } 
  previous = k
});

console.log(list[previous]);

Answer 3

Answer №2

If you're looking to find the first index in an array that is greater than a specific value, you can utilize the Array.findIndex() method. If no such index is found (resulting in a return value of -1), then the element at the last key will be selected. Otherwise, the value located at the key preceding the identified index will be chosen. In cases where the value is less than the very first key, the function will return undefined.

const findLowerClosest = (value, arr) => {
  const keys = Object.keys(list);
  const index = keys.findIndex(key => +key > value);
  const atKey = index !== -1 ? index - 1 : keys.length - 1;
  return arr[keys[atKey]];
}

const list = { 1: "a", 10: "b", 20: "c", 30: "d", 40: "e" };

console.log(findLowerClosest(15, list)); // b
console.log(findLowerClosest(75, list)); // e
console.log(findLowerClosest(-3, list)); // undefined

Answer 4

If you're looking to find the first index in an array that is greater than a specific value, you can utilize the Array.findIndex() method. If no such index is found (resulting in a return value of -1), then the element at the last key will be selected. Otherwise, the value located at the key preceding the identified index will be chosen. In cases where the value is less than the very first key, the function will return undefined.

const findLowerClosest = (value, arr) => {
  const keys = Object.keys(list);
  const index = keys.findIndex(key => +key > value);
  const atKey = index !== -1 ? index - 1 : keys.length - 1;
  return arr[keys[atKey]];
}

const list = { 1: "a", 10: "b", 20: "c", 30: "d", 40: "e" };

console.log(findLowerClosest(15, list)); // b
console.log(findLowerClosest(75, list)); // e
console.log(findLowerClosest(-3, list)); // undefined

Answer 5

Answer №3

When presented with pre-sorted keys, it is efficient to utilize a binary search algorithm with a time complexity of log(n). By arranging the keys into an array and performing a one-time linear step (with a possible one-time sort operation if needed), this method proves valuable when multiple searches on a single structure are anticipated.

const bisect = (a, target, lo=0, hi=a.length-1) => {
  while (lo <= hi) {
    const mid = ~~((hi - lo) / 2 + lo);

    if (a[mid] === target) {
      return a[mid];
    }
    else if (a[mid] < target) {
      lo = mid + 1;
    }
    else {
      hi = mid - 1;
    }
  }
  
  return a[~~((hi-lo)/2+lo)];
};

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const keys = Object.keys(list);
console.log(list[bisect(keys, 15)]);
console.log(list[bisect(keys, 16)]);
console.log(list[bisect(keys, -50)]);
console.log(list[bisect(keys, 50)]);

Answer 6

When presented with pre-sorted keys, it is efficient to utilize a binary search algorithm with a time complexity of log(n). By arranging the keys into an array and performing a one-time linear step (with a possible one-time sort operation if needed), this method proves valuable when multiple searches on a single structure are anticipated.

const bisect = (a, target, lo=0, hi=a.length-1) => {
  while (lo <= hi) {
    const mid = ~~((hi - lo) / 2 + lo);

    if (a[mid] === target) {
      return a[mid];
    }
    else if (a[mid] < target) {
      lo = mid + 1;
    }
    else {
      hi = mid - 1;
    }
  }
  
  return a[~~((hi-lo)/2+lo)];
};

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const keys = Object.keys(list);
console.log(list[bisect(keys, 15)]);
console.log(list[bisect(keys, 16)]);
console.log(list[bisect(keys, -50)]);
console.log(list[bisect(keys, 50)]);

Answer 7

Answer №4

Utilize the filter function to extract values that are below 15, then apply Math.max, and finally use this result as a key to fetch the corresponding value from the object.

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const targetValue = 15;

function retrieveValue(n) {
  let keys = Math.max.apply(null, Object.keys(list).filter((item) => {
    return item < n
  }));
  return (list[keys])
}
console.log(retrieveValue(targetValue))

Answer 8

Utilize the filter function to extract values that are below 15, then apply Math.max, and finally use this result as a key to fetch the corresponding value from the object.

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const targetValue = 15;

function retrieveValue(n) {
  let keys = Math.max.apply(null, Object.keys(list).filter((item) => {
    return item < n
  }));
  return (list[keys])
}
console.log(retrieveValue(targetValue))

Answer 9

Answer №5

One way to solve this is by first sorting the keys and then using a reduce function to determine the lower key within the target range:

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const value = 15;
const result = list[Object.keys(list)
                     .map(a => parseInt(a))
                     .sort((a, b) => a - b)
                     .reduce((acc, curr) => (acc < value && value < curr) ? acc : curr)];
                     
console.log(result);

Answer 10

One way to solve this is by first sorting the keys and then using a reduce function to determine the lower key within the target range:

const list = {
  1: "a",
  10: "b",
  20: "c",
  30: "d",
  40: "e"
};

const value = 15;
const result = list[Object.keys(list)
                     .map(a => parseInt(a))
                     .sort((a, b) => a - b)
                     .reduce((acc, curr) => (acc < value && value < curr) ? acc : curr)];
                     
console.log(result);

Determine the minimum value of an object's keys by comparing them to a given number

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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