What is the best way to find the index of the smallest value in an array using JavaScript?

Question

What is the best way to find the index of the smallest value in an array using JavaScript?

I recently created this function that currently outputs -1.

function sayHello() {
  let buildingLevelsArray = [11,10,10];
  var smallestIndex = buildingLevelsArray.indexOf(Math.max(buildingLevelsArray));
  console.log(smallestIndex);
}
sayHello();

Instead of -1, I was expecting it to be 0. Any suggestions on how to fix this issue?

javascript arrays min

Answer 1

Answer №1

You must ensure that the array is spread out to obtain the maximum value. Otherwise, if you try to access it directly as a stringed array, you will get NaN, which is not part of the array and cannot be searched.

When you spread the array, all its elements are used as parameters for the function (see spread syntax ...).

In this scenario, the operation looks like this:

Math.max(...[11, 10, 10])

will be computed as

Math.max(11, 10, 10)

function sayHello() {
  arrayEdificiosNiveis = [11, 10, 10];
  var indexMenor = arrayEdificiosNiveis.indexOf(Math.max(...arrayEdificiosNiveis));
  
  console.log(indexMenor);
}

sayHello();

A concise solution using a single loop:

However, why not utilize

v > a[r] ? i : r

(which seems more intuitive) instead of

v <= a[r] ? r : i

The issue arises from the initial comparison with the element at index zero. At that point, r = -1 and the element is a[r] = undefined.

Comparing with undefined using a relational operator such as <, <=, >, or >= always returns

false</code, resulting in the incorrect index of <code>-1

instead of zero. This error persists for every other element in the array.

const
    getFirstIndexOfMaxValue = array =>
        array.reduce((r, v, i, a) => v <= a[r] ? r : i, -1);            

console.log(getFirstIndexOfMaxValue([]));
console.log(getFirstIndexOfMaxValue([11]));
console.log(getFirstIndexOfMaxValue([11, 10]));
console.log(getFirstIndexOfMaxValue([11, 10, 10]));
console.log(getFirstIndexOfMaxValue([10, 11, 10]));
console.log(getFirstIndexOfMaxValue([10, 11, 10, 11]));

Answer 2

You must ensure that the array is spread out to obtain the maximum value. Otherwise, if you try to access it directly as a stringed array, you will get NaN, which is not part of the array and cannot be searched.

When you spread the array, all its elements are used as parameters for the function (see spread syntax ...).

In this scenario, the operation looks like this:

Math.max(...[11, 10, 10])

will be computed as

Math.max(11, 10, 10)

function sayHello() {
  arrayEdificiosNiveis = [11, 10, 10];
  var indexMenor = arrayEdificiosNiveis.indexOf(Math.max(...arrayEdificiosNiveis));
  
  console.log(indexMenor);
}

sayHello();

A concise solution using a single loop:

However, why not utilize

v > a[r] ? i : r

(which seems more intuitive) instead of

v <= a[r] ? r : i

The issue arises from the initial comparison with the element at index zero. At that point, r = -1 and the element is a[r] = undefined.

Comparing with undefined using a relational operator such as <, <=, >, or >= always returns

false</code, resulting in the incorrect index of <code>-1

instead of zero. This error persists for every other element in the array.

const
    getFirstIndexOfMaxValue = array =>
        array.reduce((r, v, i, a) => v <= a[r] ? r : i, -1);            

console.log(getFirstIndexOfMaxValue([]));
console.log(getFirstIndexOfMaxValue([11]));
console.log(getFirstIndexOfMaxValue([11, 10]));
console.log(getFirstIndexOfMaxValue([11, 10, 10]));
console.log(getFirstIndexOfMaxValue([10, 11, 10]));
console.log(getFirstIndexOfMaxValue([10, 11, 10, 11]));

Answer 3

Answer №2

The issue with the method you are using lies in the array that is being passed to the Math.max function:

const buildingLevelsArray = [11, 10, 10],
  maxLevel = Math.max(...buildingLevelsArray);
  
console.log(buildingLevelsArray.findIndex(elem => elem === maxLevel);

If you are unable to utilize the Spread syntax, an alternative approach is to use the apply function with an array as parameters for the Math.max function.

function findMax() {
  const buildingLevelsArray = [11, 10, 10],
    maxLevel = Math.max.apply(null, buildingLevelsArray),
    indexOfMax = buildingLevelsArray.indexOf(maxLevel);
  console.log(indexOfMax);
}
findMax();

Answer 4

The issue with the method you are using lies in the array that is being passed to the Math.max function:

const buildingLevelsArray = [11, 10, 10],
  maxLevel = Math.max(...buildingLevelsArray);
  
console.log(buildingLevelsArray.findIndex(elem => elem === maxLevel);

If you are unable to utilize the Spread syntax, an alternative approach is to use the apply function with an array as parameters for the Math.max function.

function findMax() {
  const buildingLevelsArray = [11, 10, 10],
    maxLevel = Math.max.apply(null, buildingLevelsArray),
    indexOfMax = buildingLevelsArray.indexOf(maxLevel);
  console.log(indexOfMax);
}
findMax();

Answer 5

Answer №3

For expert advice, consider reaching out to Raymond Chen. Interestingly enough, the straightforward C-style approach reigns supreme in this scenario. It's not surprising that Javascript engines find C style code easy to optimize.

function findIndexOfSmallestValue(arr) {
 var lowestIndex = 0;
 for (var j = 1; j < arr.length; j++) {
  if (arr[j] < arr[lowestIndex]) lowestIndex = j;
 }
 return lowestIndex;
}

Answer 6

For expert advice, consider reaching out to Raymond Chen. Interestingly enough, the straightforward C-style approach reigns supreme in this scenario. It's not surprising that Javascript engines find C style code easy to optimize.

function findIndexOfSmallestValue(arr) {
 var lowestIndex = 0;
 for (var j = 1; j < arr.length; j++) {
  if (arr[j] < arr[lowestIndex]) lowestIndex = j;
 }
 return lowestIndex;
}

Answer 7

Answer №4

const numbers = [-1, 5, -4, 7, 10, 2, 9, 6, 8];
let output = "";
let finalOutput = "";
let index = 0;

while (index <= numbers.length) {
    index++;
    let tempNum = numbers[0];
    let nextNum = numbers[index];

    if (tempNum === nextNum) {
        finalOutput = tempNum;
        output = nextNum;
    } else if (tempNum > nextNum) {
        output = nextNum;
    } else if (output < tempNum) {
        finalOutput = output;
    }
}

if (finalOutput < output) {
    console.log(finalOutput);
} else {
    console.log(output);
}

Answer 8

const numbers = [-1, 5, -4, 7, 10, 2, 9, 6, 8];
let output = "";
let finalOutput = "";
let index = 0;

while (index <= numbers.length) {
    index++;
    let tempNum = numbers[0];
    let nextNum = numbers[index];

    if (tempNum === nextNum) {
        finalOutput = tempNum;
        output = nextNum;
    } else if (tempNum > nextNum) {
        output = nextNum;
    } else if (output < tempNum) {
        finalOutput = output;
    }
}

if (finalOutput < output) {
    console.log(finalOutput);
} else {
    console.log(output);
}

What is the best way to find the index of the smallest value in an array using JavaScript?

Answer №1

Answer №2

Answer №3

Answer №4

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