This problem can be solved using a hash map (also known as a dictionary in Python). Here is a step-by-step solution:

1. Initialize an empty hash map.
2. Loop through the array, for each element do the following:
- Check if the difference between the target and the current element exists in the hash map.
- If it exists, return the index of the current element and the value from the hash map (which is the index of the other element).
- If it doesn't exist, add the current element and its index to the hash map.
3. Continue this process until you find a pair of elements that add up to the target.

Here is the Python code for the above steps:

```python
def twoSum(nums, target):
num_map = {}
for i, num in enumerate(nums):
complement = target - num
if complement in num_map:
return [num_map[complement], i]
num_map[num] = i
```

This function takes an array of integers (`nums`) and a target integer (`target`) as input, and returns the indices of the two numbers in the array that add up to the target. The time complexity of this solution is O(n), where n is the length of the input array, because we traverse the list containing n elements only once. Each look up in the table costs only O(1) time.

Question

This problem can be solved using a hash map (also known as a dictionary in Python). Here is a step-by-step solution:

1. Initialize an empty hash map.
2. Loop through the array, for each element do the following:
   - Check if the difference between the target and the current element exists in the hash map.
   - If it exists, return the index of the current element and the value from the hash map (which is the index of the other element).
   - If it doesn't exist, add the current element and its index to the hash map.
3. Continue this process until you find a pair of elements that add up to the target.

Here is the Python code for the above steps:

```python
def twoSum(nums, target):
    num_map = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in num_map:
            return [num_map[complement], i]
        num_map[num] = i
```

This function takes an array of integers (`nums`) and a target integer (`target`) as input, and returns the indices of the two numbers in the array that add up to the target. The time complexity of this solution is O(n), where n is the length of the input array, because we traverse the list containing n elements only once. Each look up in the table costs only O(1) time.

Knowee AI · Accepted Answer

This problem can be solved using a hash map (also known as a dictionary in Python). Here is a step-by-step solution:

1. Initialize an empty hash map.
2. Loop through the array, for each element do the following:
   - Check if the difference between the target and the current element exists in the hash map.
   - If it exists, return the index of the current element and the value from the hash map (which is the index of the other element).
   - If it doesn't exist, add the current element and its index to the hash map.
3. Continue this process until you find a pair of elements that add up to the target.

Here is the Python code for the above steps:

```python
def twoSum(nums, target):
    num_map = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in num_map:
            return [num_map[complement], i]
        num_map[num] = i
```

This function takes an array of integers (`nums`) and a target integer (`target`) as input, and returns the indices of the two numbers in the array that add up to the target. The time complexity of this solution is O(n), where n is the length of the input array, because we traverse the list containing n elements only once. Each look up in the table costs only O(1) time.

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