Locate the upper and lower limits within an array

Question

Locate the upper and lower limits within an array

I'm currently working on extracting the upper and lower boundaries for a numeric value within an array.

const boundaries = [15, 30, 45, 60, 75, 90];
const age = 22;

In the given example, the expected result would be:

[15, 30]

If the value matches a boundary, it will be considered the lower value in the resulting array. If it is equal to or exceeds the maximum boundary, it will be taken as the maximum value.

Examples of possible outcomes:

15 => [15, 30]
22 => [15, 30]
30 => [30, 45]
90 => [90]

I attempted to achieve this by iterating over the array using mapping, checking if the age surpasses a boundary => and then determining the relevant boundary. Despite that, I am uncertain if this is the most effective approach.

const boundaries = [15, 30, 45, 60, 75, 90];
const age = 22;

// obtain all lower values
const allLower = boundaries.map((b) => age > b ? b : null).filter(x => x);
const lower = allLower[allLower.length - 1]; // retrieve lowest
const upper = boundaries[boundaries.indexOf(lower) + 1]; // get next

const result = [lower, upper]; // construct result

console.log(result);

Is there an alternative method that is concise / more efficient / reliable for this task?

javascript arrays

Answer 1

Answer №1

What is the reasoning behind utilizing indices in this scenario? Have you considered the possibility of the boundaries array being unsorted? Perhaps it would be more straightforward to segregate the values into allLower and allUpper (containing values below and above the threshold), and then utilize min and max on the resulting arrays instead?

Here is a sample code snippet:

const boundaries = [15, 30, 45, 60, 75, 90];
const age = 22;

const allLower = boundaries.filter(x => x < age);
const allUpper = boundaries.filter(x => x > age);

const lowerBound = Math.max(...allLower);
const upperBound = Math.min(...allUpper);

Answer 2

What is the reasoning behind utilizing indices in this scenario? Have you considered the possibility of the boundaries array being unsorted? Perhaps it would be more straightforward to segregate the values into allLower and allUpper (containing values below and above the threshold), and then utilize min and max on the resulting arrays instead?

Here is a sample code snippet:

const boundaries = [15, 30, 45, 60, 75, 90];
const age = 22;

const allLower = boundaries.filter(x => x < age);
const allUpper = boundaries.filter(x => x > age);

const lowerBound = Math.max(...allLower);
const upperBound = Math.min(...allUpper);

Answer 3

Answer №2

Using the reduce method seems like a suitable solution in this scenario:

const boundaries = [15, 30, 45, 60, 75, 90];


for (let search of [1, 22, 30, 90, 100]) {

    let [low, upr] = boundaries.reduce(([low, upr], x) =>
        [
            x <= search ? Math.max(low, x) : low,
            x >  search ? Math.min(upr, x) : upr,

        ],
        [-Infinity, +Infinity]
    )

    console.log(low, '<=', search, '<', upr)

}

This particular approach does not necessitate the sorting of boundaries. However, if they are always sorted, you may want to consider implementing binary search to find the lower bound.

Answer 4

Using the reduce method seems like a suitable solution in this scenario:

const boundaries = [15, 30, 45, 60, 75, 90];


for (let search of [1, 22, 30, 90, 100]) {

    let [low, upr] = boundaries.reduce(([low, upr], x) =>
        [
            x <= search ? Math.max(low, x) : low,
            x >  search ? Math.min(upr, x) : upr,

        ],
        [-Infinity, +Infinity]
    )

    console.log(low, '<=', search, '<', upr)

}

This particular approach does not necessitate the sorting of boundaries. However, if they are always sorted, you may want to consider implementing binary search to find the lower bound.

Answer 5

Answer №3

Perhaps considering a simple for-loop can be beneficial in this situation ;)

function determineAgeBounds(age) {
  for (let index = 0; index < ageBoundaries.length; index++) {
    if (ageBoundaries[index] <= age && (ageBoundaries[index + 1] ?? Infinity) > age) {
      return ageBoundaries.slice(index, index + 2);
    }
  }
}

Answer 6

Perhaps considering a simple for-loop can be beneficial in this situation ;)

function determineAgeBounds(age) {
  for (let index = 0; index < ageBoundaries.length; index++) {
    if (ageBoundaries[index] <= age && (ageBoundaries[index + 1] ?? Infinity) > age) {
      return ageBoundaries.slice(index, index + 2);
    }
  }
}

Answer 7

Answer №4

To identify the previous and next values, you can apply a filter function to the array.

const
    findNeighbours = (arr, middle) => arr
        .filter((val, index, { [index - 1]: prev, [index + 1]: next }) =>
            val <= middle && next > middle ||
            prev <= middle && val >= middle ||
            prev === undefined && next > middle ||
            prev < middle && next === undefined
        ),
    limits = [18, 36, 54, 72, 90];

console.log(...findNeighbours(limits, 22));  // in between
console.log(...findNeighbours(limits, 36));  // directly above and below
console.log(...findNeighbours(limits, 10));  // lowest value 
console.log(...findNeighbours(limits, 18));  // directly above and below
console.log(...findNeighbours(limits, 90));  // highest value
console.log(...findNeighbours(limits, 100)); // highest value

Answer 8

To identify the previous and next values, you can apply a filter function to the array.

const
    findNeighbours = (arr, middle) => arr
        .filter((val, index, { [index - 1]: prev, [index + 1]: next }) =>
            val <= middle && next > middle ||
            prev <= middle && val >= middle ||
            prev === undefined && next > middle ||
            prev < middle && next === undefined
        ),
    limits = [18, 36, 54, 72, 90];

console.log(...findNeighbours(limits, 22));  // in between
console.log(...findNeighbours(limits, 36));  // directly above and below
console.log(...findNeighbours(limits, 10));  // lowest value 
console.log(...findNeighbours(limits, 18));  // directly above and below
console.log(...findNeighbours(limits, 90));  // highest value
console.log(...findNeighbours(limits, 100)); // highest value

Answer 9

Answer №5

To simplify the array, you can do something along these lines:

const boundaries = [15, 30, 45, 60, 75, 90];

const getResult = (array, target) => {
  if (target < array[0] || target > array[array.length - 1]) {
    return [];
  }
  return array.reduce((a, c) => {
    if (c <= target) {
      a[0] = c;
    } else if (c > target && (!a[1] || c < a[a.length - 1])) {
      a[a.length] = c;
    }
    return a;
  }, []);
}

console.log(getResult(boundaries, 22));
console.log(getResult(boundaries, 15));
console.log(getResult(boundaries, 30));
console.log(getResult(boundaries, 90));
console.log(getResult(boundaries, 14));
console.log(getResult(boundaries, 91));

Answer 10

To simplify the array, you can do something along these lines:

const boundaries = [15, 30, 45, 60, 75, 90];

const getResult = (array, target) => {
  if (target < array[0] || target > array[array.length - 1]) {
    return [];
  }
  return array.reduce((a, c) => {
    if (c <= target) {
      a[0] = c;
    } else if (c > target && (!a[1] || c < a[a.length - 1])) {
      a[a.length] = c;
    }
    return a;
  }, []);
}

console.log(getResult(boundaries, 22));
console.log(getResult(boundaries, 15));
console.log(getResult(boundaries, 30));
console.log(getResult(boundaries, 90));
console.log(getResult(boundaries, 14));
console.log(getResult(boundaries, 91));

Locate the upper and lower limits within an array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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