return an array of indices of minimum numbers in a array

Question

We have an array of numbers like this:

const array = [2, 1, 3, 1];

the minimum number in the array is 1 right? Now I want to return a new array containing the index or indices of the minimum numbers in the given array. So in above array the desired result would be:

[1, 3] // indices of minimum number which is 1

If there is only one element in the array we should return 0 right...

Please include your code that attempts to implement this and explain how it's failing so someone can help you. — jarmod, Jan 24 '21 at 20:03
There are many good solutions here with explanations: [Smallest number in array and its position](https://stackoverflow.com/questions/15455247/smallest-number-in-array-and-its-position) — Zac Anger, Jan 24 '21 at 20:07

score 1 · Answer 1 · answered Jan 24 '21 at 20:03

You could use indexOf and a while loop:

const array = [2, 1, 3, 1];
const minElement = Math.min(...array);
const minElementIndexes = [];
let indexOfMinElement = array.indexOf(minElement);
while (indexOfMinElement != -1) {
    minElementIndexes.push(indexOfMinElement);
    indexOfMinElement = array.indexOf(minElement, indexOfMinElement + 1);
}
console.log(minElementIndexes);

score 1 · Accepted Answer · answered Jan 24 '21 at 20:04

Step by step:

Determine the minimum number.
Map the array so that the minimum numbers are replaced with the indexes, and other numbers are replaced with undefineds.
Filter out the undefineds.

const array = [2, 1, 3, 1]
const min = Math.min(...array) // 1
const result = array
  .map((number, index) => number === min ? index : undefined) // 2
  .filter((item) => item !== undefined) // 3
console.log(result)

score 1 · Answer 3 · answered Jan 24 '21 at 20:05

Get the min number, then iterates through the array by his index:

const array = [2, 1, 3, 1];

let reduced = array.reduce((x, y) => x < y ? x : y);
console.log(reduced)
let result = [];

for(let i in array){
    if(array[i] == reduced){
        result.push(i)
    }
}
console.log(result)

buzatto · Answer 4 · 2021-01-24T20:43:19.610

0

you can use one loop where you keep tracking minValue:

const array = [2, 1, 3, 1];

let minValue = array[0]; 

const indexes = array.reduce((arr, val, idx) => {
  if (idx === 0) return [idx];
  if (val > minValue) return arr;
  if (val === minValue) {
    arr.push(idx);
    return arr;
  };
  minValue = val;
  return [idx];
}, []);

console.log(indexes)

edited Jan 24 '21 at 20:43

answered Jan 24 '21 at 20:35

buzatto

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return an array of indices of minimum numbers in a array

4 Answers4