Locating the index of a specific number within an array

Question

Locating the index of a specific number within an array

Given an array of numbers like this:

let numbers = [4,7,2,0,9];

I would like to map the numbers in the array based on their size and return their positions. The expected output should be:

numbers = [3,4,2,1,5];

Thank you!

javascript arrays sorting

Answer 1

Answer №1

This solution is the most straightforward one among all the answers provided

For smaller arrays, this method will work effectively. However, for larger arrays, it is recommended to use a lookup Map for better performance.

const arr = [4,7,2,0,9]

const indices = arr.slice(0).sort((a,b) => a-b); // make a sorted copy of array
const result = arr.map(item => indices.indexOf(item)+1); // map indices plus 1

console.log(result)

Answer 2

This solution is the most straightforward one among all the answers provided

For smaller arrays, this method will work effectively. However, for larger arrays, it is recommended to use a lookup Map for better performance.

const arr = [4,7,2,0,9]

const indices = arr.slice(0).sort((a,b) => a-b); // make a sorted copy of array
const result = arr.map(item => indices.indexOf(item)+1); // map indices plus 1

console.log(result)

Answer 3

Answer №2

Replicate the array, arrange it in ascending order, and pair each number with its index + 1. Transform the pairs into an object called position.

Next, iterate through the original array and fetch the position of each number from the positions object.

const numbers = [4,7,2,0,9];

const positions = Object.fromEntries(
  [...numbers]
    .sort((a, b) => a - b)
    .map((num, index) => [num, index + 1])
);

const result = numbers.map(num => positions[num]);

console.log(result);

You can utilize a Map instead of an object by substituting Object.fromEntries() with the constructor of Map:

const numbers = [4,7,2,0,9];

const positions = new Map(
  [...numbers]
    .sort((a, b) => a - b)
    .map((num, index) => [num, index + 1])
);

const result = numbers.map(num => positions.get(num));

console.log(result);

To handle numerous identical values, condense the array into a Map and incorporate the current index. While iterating through the initial array, extract the index of the value from the beginning of each entry's value:

const numbers = [1, 2, 1, 0, 0, 0];

const positions = [...numbers]
  .sort((a, b) => a - b)
  .reduce((acc, v, i) => acc.set(v, [...acc.get(v) ?? [], i + 1]), new Map())

const result = numbers.map(num => positions.get(num).shift());

console.log(result);

Answer 4

Replicate the array, arrange it in ascending order, and pair each number with its index + 1. Transform the pairs into an object called position.

Next, iterate through the original array and fetch the position of each number from the positions object.

const numbers = [4,7,2,0,9];

const positions = Object.fromEntries(
  [...numbers]
    .sort((a, b) => a - b)
    .map((num, index) => [num, index + 1])
);

const result = numbers.map(num => positions[num]);

console.log(result);

You can utilize a Map instead of an object by substituting Object.fromEntries() with the constructor of Map:

const numbers = [4,7,2,0,9];

const positions = new Map(
  [...numbers]
    .sort((a, b) => a - b)
    .map((num, index) => [num, index + 1])
);

const result = numbers.map(num => positions.get(num));

console.log(result);

To handle numerous identical values, condense the array into a Map and incorporate the current index. While iterating through the initial array, extract the index of the value from the beginning of each entry's value:

const numbers = [1, 2, 1, 0, 0, 0];

const positions = [...numbers]
  .sort((a, b) => a - b)
  .reduce((acc, v, i) => acc.set(v, [...acc.get(v) ?? [], i + 1]), new Map())

const result = numbers.map(num => positions.get(num).shift());

console.log(result);

Locating the index of a specific number within an array

Answer №1

Answer №2

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