Sure, here is a Python solution for your problem using a dictionary (hashmap) to store the numbers and their indices.

```python
def twoSum(nums, target):
if len(nums)

Question

Sure, here is a Python solution for your problem using a dictionary (hashmap) to store the numbers and their indices.

```python
def twoSum(nums, target):
    if len(nums) <= 1:
        return False
    buff_dict = {}
    for i in range(len(nums)):
        if nums[i] in buff_dict:
            return [buff_dict[nums[i]], i]
        else:
            buff_dict[target - nums[i]] = i
```

Here is a step by step explanation:

1. Check if the length of the input list is less than or equal to 1. If it is, return False because there are not enough numbers to add up to the target.

2. Create an empty dictionary `buff_dict`.

3. Loop through the list `nums` with the index `i`.

4. Check if `nums[i]` is in `buff_dict`. If it is, return a list with `buff_dict[nums[i]]` and `i`. These are the indices of the two numbers that add up to the target.

5. If `nums[i]` is not in `buff_dict`, add a new entry to the dictionary with the key as `target - nums[i]` and the value as `i`. This is because if we subtract `nums[i]` from the target, we get the other number that we need to find in the list.

6. Continue this process until we find a pair of numbers that add up to the target or we have checked all the numbers in the list.

This solution has a time complexity of O(n) because we are making a single pass through the list, and a space complexity of O(n) because in the worst case, we will insert every element of the list into the dictionary.

Knowee AI · Accepted Answer

Sure, here is a Python solution for your problem using a dictionary (hashmap) to store the numbers and their indices.

```python
def twoSum(nums, target):
    if len(nums) <= 1:
        return False
    buff_dict = {}
    for i in range(len(nums)):
        if nums[i] in buff_dict:
            return [buff_dict[nums[i]], i]
        else:
            buff_dict[target - nums[i]] = i
```

Here is a step by step explanation:

1. Check if the length of the input list is less than or equal to 1. If it is, return False because there are not enough numbers to add up to the target.

2. Create an empty dictionary `buff_dict`.

3. Loop through the list `nums` with the index `i`.

4. Check if `nums[i]` is in `buff_dict`. If it is, return a list with `buff_dict[nums[i]]` and `i`. These are the indices of the two numbers that add up to the target.

5. If `nums[i]` is not in `buff_dict`, add a new entry to the dictionary with the key as `target - nums[i]` and the value as `i`. This is because if we subtract `nums[i]` from the target, we get the other number that we need to find in the list.

6. Continue this process until we find a pair of numbers that add up to the target or we have checked all the numbers in the list.

This solution has a time complexity of O(n) because we are making a single pass through the list, and a space complexity of O(n) because in the worst case, we will insert every element of the list into the dictionary.

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

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