Locating the index of the initial duplicate in an array using Javascript

Question

Locating the index of the initial duplicate in an array using Javascript


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

I’m looking to write a function called indexOfRepeatedValue (array) that utilizes the numbers stored in the provided variable. The function should create a variable called firstIndex. Within a for loop, determine which number repeats first and assign its index to firstIndex. Finally, log this variable to the console beyond the for loop.

This initial idea isn’t working as intended. I could use some guidance or advice on how to proceed. Any help would be greatly appreciated!

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

function indexOfRepeatedValue(array) {
  let firstIndex;
  for (let i = 0; i < array.length; i++)
    if (firstIndex.indexOf(array[i]) === -1 && array[i] !== '');
  firstIndex.push(array[i]);
  return firstIndex;
}

console.log(
  indexOfRepeatedValue(numbers)
)

javascript arrays function duplicates

Answer 1

Answer №1

To begin, create an array called firstIndex: let firstIndex = [];

Ensure that the variable i is within the scope you have defined using let.

Remember to end each statement with a semicolon to prevent the loop from moving to the next line prematurely.

Then, find and return the first number in the new array that matches another number.

Keep in mind that in JavaScript arrays start at index 0, so the result will be 3 because the second occurrence of number 2 is in the 4th position.

I have tried to maintain the structure of your code as closely as possible.

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

function indexOfRepeatedValue(array) {
  let firstIndex = [];
  for (let i = 0; i < array.length; i++) {
    if (firstIndex.indexOf(array[i]) !== -1) { // duplicate found
      console.log("found",array[i], "again at index", i)
      console.log("The first occurrence of this value is at index",numbers.indexOf(array[i]))
      
      return i; // stop and return once found
      // use `return numbers.indexOf(array[i])` if you want the index of the first duplicate   
    }
    firstIndex.push(array[i]); // no duplicate found yet
  }
  return "no duplicates found"
}

console.log(
  indexOfRepeatedValue(numbers)
)

There are multiple approaches to solving this problem.

JavaScript: How to find first duplicate value and return its index?

Answer 2

To begin, create an array called firstIndex: let firstIndex = [];

Ensure that the variable i is within the scope you have defined using let.

Remember to end each statement with a semicolon to prevent the loop from moving to the next line prematurely.

Then, find and return the first number in the new array that matches another number.

Keep in mind that in JavaScript arrays start at index 0, so the result will be 3 because the second occurrence of number 2 is in the 4th position.

I have tried to maintain the structure of your code as closely as possible.

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

function indexOfRepeatedValue(array) {
  let firstIndex = [];
  for (let i = 0; i < array.length; i++) {
    if (firstIndex.indexOf(array[i]) !== -1) { // duplicate found
      console.log("found",array[i], "again at index", i)
      console.log("The first occurrence of this value is at index",numbers.indexOf(array[i]))
      
      return i; // stop and return once found
      // use `return numbers.indexOf(array[i])` if you want the index of the first duplicate   
    }
    firstIndex.push(array[i]); // no duplicate found yet
  }
  return "no duplicates found"
}

console.log(
  indexOfRepeatedValue(numbers)
)

There are multiple approaches to solving this problem.

JavaScript: How to find first duplicate value and return its index?

Answer 3

Answer №2

To efficiently find the index of a repeated value in an array, you can create an object to store the index of each element and return early if the index already exists.

function findIndexOfRepeatedValue(arr) {
    let idx = {};
    for (let i = 0; i < arr.length; i++) {
        if (idx[arr[i]] !== undefined) return idx[arr[i]];
        idx[arr[i]] = i;
    }
    return -1;
}

const elements = [2, 4, 5, 2, 3, 5, 1, 2, 4];
console.log(findIndexOfRepeatedValue(elements));

Answer 4

To efficiently find the index of a repeated value in an array, you can create an object to store the index of each element and return early if the index already exists.

function findIndexOfRepeatedValue(arr) {
    let idx = {};
    for (let i = 0; i < arr.length; i++) {
        if (idx[arr[i]] !== undefined) return idx[arr[i]];
        idx[arr[i]] = i;
    }
    return -1;
}

const elements = [2, 4, 5, 2, 3, 5, 1, 2, 4];
console.log(findIndexOfRepeatedValue(elements));

Answer 5

Answer №3

To begin, start by setting up the variable firstIndex:

let firstIndex = [];

The next step is to locate the index of each repeated element using the following code:

if( array.slice(0,i).includes(array[i]) ) {
    firstIndex.push( i );
}

If you are looking for the absolute first index of a repeat, simply return:

return firstIndex[0];
//Keep in mind that if your goal is solely to find the first index, there is no need for the firstIndex variable or to iterate through the entire loop.

If you require all indices of repeated elements:

return firstIndex;

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

function indexOfRepeatedValue(array) {
  let firstIndex = [];
  for (let i = 0; i < array.length; i++)
    if( array.slice(0,i).includes(array[i]) ) {
      firstIndex.push(i);
    }
  return firstIndex[0];
}

console.log(
  indexOfRepeatedValue(numbers)
)

NOTE

Another approach is to utilize Array#map to obtain the index of repeated values and then use Array#filter to retain only those indices, where the first instance can be accessed with [0].

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

const indexOfRepeatedValue = arr => 
    arr.map((a,i) => arr.slice(0,i).includes(a) ? i : -1)
    .filter(i => i > -1)[0];
    
console.log( indexOfRepeatedValue( numbers ) );

Answer 6

To begin, start by setting up the variable firstIndex:

let firstIndex = [];

The next step is to locate the index of each repeated element using the following code:

if( array.slice(0,i).includes(array[i]) ) {
    firstIndex.push( i );
}

If you are looking for the absolute first index of a repeat, simply return:

return firstIndex[0];
//Keep in mind that if your goal is solely to find the first index, there is no need for the firstIndex variable or to iterate through the entire loop.

If you require all indices of repeated elements:

return firstIndex;

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

function indexOfRepeatedValue(array) {
  let firstIndex = [];
  for (let i = 0; i < array.length; i++)
    if( array.slice(0,i).includes(array[i]) ) {
      firstIndex.push(i);
    }
  return firstIndex[0];
}

console.log(
  indexOfRepeatedValue(numbers)
)

NOTE

Another approach is to utilize Array#map to obtain the index of repeated values and then use Array#filter to retain only those indices, where the first instance can be accessed with [0].

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

const indexOfRepeatedValue = arr => 
    arr.map((a,i) => arr.slice(0,i).includes(a) ? i : -1)
    .filter(i => i > -1)[0];
    
console.log( indexOfRepeatedValue( numbers ) );

Locating the index of the initial duplicate in an array using Javascript

Answer №1

Answer №2

Answer №3

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