I walk you through how to solve Leetcode problem <a target="_blank" href="https://leetcode.com/problems/valid-anagram/description/">1.</a> Two Sum. I highlight a couple of approaches to solving it. As usual, there are various trade-offs between time and space complexity respective to each approach.
<h2 id="heading-approach-1-brute-force">Approach 1: Brute Force</h2>
<pre><code class="lang-python">def twoSum(nums, target):
 for i in range(len(nums) - 1):
 for j in range(i + 1, len(nums)):
 if nums[i] + nums[j] == target:
 return [i,j]
</code></pre>
Time Complexity: O(N²) Space Complexity: O(1)
<h2 id="heading-approach-2-hashmap">Approach 2: HashMap</h2>
<pre><code class="lang-python">def twoSum(nums, target):
 numbersSeen = {}
 for index, num in enumerate(nums):
 diff = target - nums[index]
 if diff in numbersSeen:
 return [numbersSeen[diff], index]
 numbersSeen[nums[index]] = index
</code></pre>
Time Complexity: O(N) Space Complexity: O(N)

I walk you through how to solve Leetcode problem [**1\.**](https://leetcode.com/problems/valid-anagram/description/) **Two Sum**. I highlight a couple of approaches to solving it. As usual, there are various trade-offs between time and space complexity respective to each approach.

## **Approach 1: Brute Force**

```python
def twoSum(nums, target):
    for i in range(len(nums) - 1):
        for j in range(i + 1, len(nums)):
            if nums[i] + nums[j] == target:
                return [i,j]
```

**Time Complexity: O(N²**)  
**Space Complexity: O(1)**

## **Approach 2: HashMap**

```python
def twoSum(nums, target):
        numbersSeen = {}
        for index, num in enumerate(nums):
            diff = target - nums[index]
            if diff in numbersSeen:
                return [numbersSeen[diff], index]
            numbersSeen[nums[index]] = index
```

**Time Complexity: O(N**)  
**Space Complexity: O(N)**

03-1 : Two Sum

Arrays / Hashing

Approach 1: Brute Force

Approach 2: HashMap