Understanding the Effect of delete on Arrays in JavaScript? – EPAM Interview Insight

Step-by-Step Explanation:

Step 1: Declare the Array

let nums = [1, 2, 3, 4];

Now nums is:

Index:  0   1   2   3
Value: [1,  2,  3,  4]

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Step 2: Use the delete Operator

delete nums[2];

This line removes the value at index 2, but it does NOT reindex or shorten the array.

Important:

• delete removes the value but leaves a hole (called a "sparse array").
• It doesn’t affect the array’s length.

So after deletion:

nums = [1, 2, empty, 4];

Internally:

Index:  0    1    2     3
Value: [1,   2,  empty, 4]

The value at index 2 is now undefined, but not explicitly. It's a missing slot — not the same as undefined.

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NOTE :- Check the Length

console.log(nums.length); // 4

Yup! Even after deleting an item, the array length stays the same.

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Step 3: Log the Array

console.log(nums);
// Output: [1, 2, empty, 4]

Depending on your environment:

• Some will show: [1, 2, , 4]
• Some (like Node.js) might show: [1, 2, <1 empty item>, 4]

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✔ Final Answer:

D) [1,2,empty,4]

This question appeared in the Epam 2nd round of interview, testing the candidate’s knowledge of use of delete keyword with array in JavaScript.

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Summary

Concept	Behavior
delete nums[2]	Removes the value at index 2
Array stays same	Length remains 4
Index 2 becomes A "hole"	(not undefined)

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When to Avoid delete in Arrays

delete is not the recommended way to remove elements from arrays.

Instead, use:

• .splice() removes and reindexes:

    nums.splice(2, 1); // removes 1 element at index 2

    ➤ Result: [1, 2, 4] and length becomes 3