Ways to develop a function that provides the index of the initial duplicate value within an array

Question

Ways to develop a function that provides the index of the initial duplicate value within an array

My task is:

To develop a function named, indexOfRepeatedValue(arr),
To establish a firstIndex variable within the function,
To utilize a for loop to identify the first repeated number in a given array,
And to store the index of that number in the created firstIndex variable

The provided array is as follows:

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

For this array, the first replicated number is 2. Therefore, the value of the firstIndex variable should be 0 (upon returning the value from the function).

My code progression is shown below,

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

function indexOfRepeatedValue(arr) {
    let firstIndex;
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] === ) { //The area where I am missing a crucial element
            firstIndex = ;
        }
        break;
    }
    return firstIndex;
}

console.log(indexOfRepeatedValue(givenArr));

I am uncertain of the correct value to include in the if statement. I am unable to use the indexOf function at this time. Any guidance would be greatly appreciated as I am eager to learn.

javascript arrays loops for-loop

Answer 1

Answer №1

To identify duplicates in an array, a Set can be used. If the loop comes across an element that is already in the set, it means a match has been found and the firstIndex variable can be set.

Keep in mind: The value of firstIndex will be undefined if the array does not contain any duplicates.

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

function findDuplicateIndex(arr) {
    let firstIndex;
    let s = new Set();
    for (let i = 0; i < arr.length; i++) {
        if (s.has(arr[i])) { 
            firstIndex = arr.indexOf(arr[i]);
            break;
        } else {
          s.add(arr[i])
        }
    }
    return firstIndex;
}

console.log(findDuplicateIndex(givenArr));

Considering the suggestion by the original poster

If you prefer not to use an additional data structure, you can instead check if the current element has been encountered before. However, it is recommended to use the initial method if possible.

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

function findDuplicateIndex(arr) {
    let firstIndex;
    for (let i = 0; i < arr.length; i++) {
        if (arr.indexOf(arr[i]) < i) { 
            firstIndex = arr.indexOf(arr[i]);
            break;
        }
    }
    return firstIndex;
}

console.log(findDuplicateIndex(givenArr));

Answer 2

To identify duplicates in an array, a Set can be used. If the loop comes across an element that is already in the set, it means a match has been found and the firstIndex variable can be set.

Keep in mind: The value of firstIndex will be undefined if the array does not contain any duplicates.

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

function findDuplicateIndex(arr) {
    let firstIndex;
    let s = new Set();
    for (let i = 0; i < arr.length; i++) {
        if (s.has(arr[i])) { 
            firstIndex = arr.indexOf(arr[i]);
            break;
        } else {
          s.add(arr[i])
        }
    }
    return firstIndex;
}

console.log(findDuplicateIndex(givenArr));

Considering the suggestion by the original poster

If you prefer not to use an additional data structure, you can instead check if the current element has been encountered before. However, it is recommended to use the initial method if possible.

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

function findDuplicateIndex(arr) {
    let firstIndex;
    for (let i = 0; i < arr.length; i++) {
        if (arr.indexOf(arr[i]) < i) { 
            firstIndex = arr.indexOf(arr[i]);
            break;
        }
    }
    return firstIndex;
}

console.log(findDuplicateIndex(givenArr));

Answer 3

Answer №2

If you want to find the first index of a repeated value in an array, you can create a function that stores the indices of the values it encounters and checks if a value has been seen before. If a repeated value is found, the function will return the index of the first occurrence.

function findFirstRepeatedIndex(arr) {
    const indices = {};
    for (let i = 0; i < arr.length; i++) {
        const value = arr[i];
        if (value in indices) return indices[value];
        indices[value] = i;
    }
    return -1; // Return -1 if no repeated value is found
}

console.log(findFirstRepeatedIndex([7, 13, 2, 4, 5, 2, 3, 5, 1, 2, 4]));

Answer 4

If you want to find the first index of a repeated value in an array, you can create a function that stores the indices of the values it encounters and checks if a value has been seen before. If a repeated value is found, the function will return the index of the first occurrence.

function findFirstRepeatedIndex(arr) {
    const indices = {};
    for (let i = 0; i < arr.length; i++) {
        const value = arr[i];
        if (value in indices) return indices[value];
        indices[value] = i;
    }
    return -1; // Return -1 if no repeated value is found
}

console.log(findFirstRepeatedIndex([7, 13, 2, 4, 5, 2, 3, 5, 1, 2, 4]));

Answer 5

Answer №3

concise

const findRepeatedIndex = a => [...a].findIndex(function(x) {
  return (l = this.l = this.l || {}) && l[x] && 1 || (l[x] = 1) && 0;
});
console.log(findRepeatedIndex([2, 4, 5, 2, 3, 5, 1, 2, 4]));

skip array mutation by using spread syntax
findIndex locates the first index with a truthy value or returns -1 if not found
a function is essential for maintaining the correct this context, unlike arrow syntax
utilizing this allows the addition of l to the array to monitor values
the return statement evaluates as truthy if l[x] is equal to 1
otherwise, (l[x] = 1) && 0 combines assignment with a falsy condition

Answer 6

concise

const findRepeatedIndex = a => [...a].findIndex(function(x) {
  return (l = this.l = this.l || {}) && l[x] && 1 || (l[x] = 1) && 0;
});
console.log(findRepeatedIndex([2, 4, 5, 2, 3, 5, 1, 2, 4]));

skip array mutation by using spread syntax
findIndex locates the first index with a truthy value or returns -1 if not found
a function is essential for maintaining the correct this context, unlike arrow syntax
utilizing this allows the addition of l to the array to monitor values
the return statement evaluates as truthy if l[x] is equal to 1
otherwise, (l[x] = 1) && 0 combines assignment with a falsy condition

Ways to develop a function that provides the index of the initial duplicate value within an array

Answer №1

Answer №2

Answer №3

concise

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