** Problem Solving - Sum of Two Numbers (Hash Table)**

The `twoSum`

function is designed to find two numbers in an array (`numbers`

) that add up to a specific target value. The function returns the indices of these two numbers in the form of an array. Here's a step-by-step explanation:

function twoSum(numbers = [], target = 0) {

const hashTable = {};for (let i = 0; i < numbers.length; i++) {const remaining = target - nums[i]if (hashTable[remaining] !== undefined) {return [hashTable[remaining], i]}hashTable[numbers[i]] = i;}return [];};

### Function Breakdown:

**Input Parameters:**`nums`

: An array of numbers.`target`

: A single number representing the target sum.

**Output:**- An array containing the indices of the two numbers that add up to the
`target`

. If no such pair is found, the function returns an empty array.

- An array containing the indices of the two numbers that add up to the

### Explanation:

Initialization:

A hash table (or object)

`hashTable`

is initialized to store the indices of numbers encountered so far.A loop is set up to iterate over the

`nums`

array, where `i`

is the current index.For each number in the array, the function calculates the

`remaining`

value that, when added to `nums[i]`

, would equal the `target`

.`remaining`

value already exists in the `hashTable`

.`target`

. The function then returns an array with the index of the `remaining`

value and the current index `i`

.`remaining`

value is not found in the hash table, the current number (`nums[i]`

) and its index (`i`

) are stored in the `hashTable`

.`target`

, the function returns an empty array, indicating that no such pair exists.### Example:

Suppose

`nums = [2, 7, 11, 15]`

and`target = 9`

.

Iteration 1:`i = 0`

,`nums[0] = 2`

,`remaining = 9 - 2 = 7`

`hashTable = {}`

(7 is not in`hashTable`

)

`hashTable`

becomes`{2: 0}`

`i = 1`

,`nums[1] = 7`

,`remaining = 9 - 7 = 2`

`hashTable = {2: 0}`

(2 is in`hashTable`

)

- The function returns
`[0, 1]`

since`nums[0] + nums[1] = 2 + 7 = 9`

.

This solution is efficient with a time complexity of O(n), where n is the length of the `nums`

array. This is because each element is processed only once in a single pass through the array.

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