# 1. Two Sum - LeetCode Best Practices

Visit original link: [1. Two Sum - LeetCode Best Practices](https://leetcoder.net/en/leetcode/1-two-sum) for a better experience!

LeetCode link: [1. Two Sum](https://leetcode.com/problems/two-sum), difficulty: **Easy**.

## LeetCode description of "1. Two Sum"

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

You may assume that each input would have ***exactly* one solution**, and you may not use the same element twice.

You can return the answer in any order.

### [Example 1]

**Input**: `nums = [2,7,11,15], target = 9`

**Output**: `[0,1]`

**Explanation**: 

<p>Because nums[0] + nums[1] == 9, we return [0, 1].</p>


### [Example 2]

**Input**: `nums = [3,2,4], target = 6`

**Output**: `[1,2]`

### [Example 3]

**Input**: `nums = [3,3], target = 6`

**Output**: `[0,1]`

### [Constraints]

- `2 <= nums.length <= 10^4`
- `-10^9 <= nums[i] <= 10^9`
- `-10^9 <= target <= 10^9`
- **Only one valid answer exists.**

### [Hints]

<details>
  <summary>Hint 1</summary>
  A really brute force way would be to search for all possible pairs of numbers but that would be too slow. Again, it's best to try out brute force solutions for just for completeness. It is from these brute force solutions that you can come up with optimizations.

  
</details>

<details>
  <summary>Hint 2</summary>
  So, if we fix one of the numbers, say `x`, we have to scan the entire array to find the next number `y` which is `value - x` where value is the input parameter. Can we change our array somehow so that this search becomes faster?

  
</details>

<details>
  <summary>Hint 3</summary>
  The second train of thought is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?

  
</details>

## Intuition 1

1. The time complexity of the brute force solution is `O(n^2)`. To improve efficiency, you can sort the array, and then use **two pointers**, one pointing to the head of the array and the other pointing to the tail of the array, and decide `left += 1` or `right -= 1` according to the comparison of `sum` and `target`.

2. After sorting an array of numbers, if you want to know the original `index` corresponding to a certain value, there are two solutions:

<details><summary>Click to view the answer</summary><p>
    - Solution 1: Bring the `index` when sorting, that is, the object to be sorted is an array of tuples of `(num, index)`. This technique **must be mastered**, as it will be used in many questions.
    - Solution 2: Use `index()` method to find it. I have discussed this in another solution.
</p></details>

## Complexity

- Time complexity: `O(N * log N)`.
- Space complexity: `O(N)`.

## Python

```python
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        num_index_list = [(num, i) for i, num in enumerate(nums)]
        num_index_list.sort()

        left = 0
        right = len(nums) - 1

        while left < right:
            sum_ = num_index_list[left][0] + num_index_list[right][0]
            
            if sum_ == target:
                return [num_index_list[left][1], num_index_list[right][1]]

            if sum_ < target:
                left += 1
                continue

            right -= 1
```

## Other languages

```java
// Welcome to create a PR to complete the code of this language, thanks!
```

## Intuition 2

1. In `Map`, `key` is `num`, and `value` is array `index`.
2. Traverse the array, if `target - num` is in `Map`, return it. Otherwise, add `num` to `Map`.

## Steps

1. In `Map`, `key` is `num`, and `value` is array `index`.

    ```javascript
    let numToIndex = new Map()
    
    for (let i = 0; i < nums.length; i++) {
      numToIndex.set(nums[i], i)
    }
    ```

2. Traverse the array, if `target - num` is in `Map`, return it. Otherwise, add `num` to `Map`.

    ```javascript
    let numToIndex = new Map()
    
    for (let i = 0; i < nums.length; i++) {
      if (numToIndex.has(target - nums[i])) { // 1
        return [numToIndex.get(target - nums[i]), i] // 2
      }
    
      numToIndex.set(nums[i], i)
    }
    ```

## Complexity

- Time complexity: `O(N)`.
- Space complexity: `O(N)`.

## Java

```java
class Solution {
    public int[] twoSum(int[] nums, int target) {
        var numToIndex = new HashMap<Integer, Integer>();
  
        for (var i = 0; i < nums.length; i++) {
            if (numToIndex.containsKey(target - nums[i])) {
                return new int[]{numToIndex.get(target - nums[i]), i};
            }

            numToIndex.put(nums[i], i);
        }

        return null;
    }
}
```

## Python

```python
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        num_to_index = {}
        
        for i, num in enumerate(nums):
            if target - num in num_to_index:
                return [num_to_index[target - num], i]

            num_to_index[num] = i
```

## C++

```cpp
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> num_to_index;
  
        for (auto i = 0; i < nums.size(); i++) {
            if (num_to_index.contains(target - nums[i])) {
                return {num_to_index[target - nums[i]], i};
            }

            num_to_index[nums[i]] = i;
        }

        return {};
    }
};
```

## JavaScript

```javascript
var twoSum = function (nums, target) {
  let numToIndex = new Map()
  
  for (let i = 0; i < nums.length; i++) {
    if (numToIndex.has(target - nums[i])) {
      return [numToIndex.get(target - nums[i]), i]
    }

    numToIndex.set(nums[i], i)
  }
};
```

## C#

```csharp
public class Solution {
    public int[] TwoSum(int[] nums, int target) {
        var numToIndex = new Dictionary<int, int>();
  
        for (int i = 0; i < nums.Length; i++) {
            if (numToIndex.ContainsKey(target - nums[i])) {
                return [numToIndex[target - nums[i]], i];
            }

            numToIndex[nums[i]] = i;
        }

        return null;
    }
}
```

## Go

```go
func twoSum(nums []int, target int) []int {
    numToIndex := map[int]int{}

    for i, num := range nums {
        if index, ok := numToIndex[target - num]; ok {
            return []int{index, i}
        }

        numToIndex[num] = i
    }

    return nil
}
```

## Ruby

```ruby
def two_sum(nums, target)
  num_to_index = {}

  nums.each_with_index do |num, i|
    if num_to_index.key?(target - num)
      return [num_to_index[target - num], i]
    end

    num_to_index[num] = i
  end
end
```

## Other languages

```java
// Welcome to create a PR to complete the code of this language, thanks!
```

## Intuition 3

1. The time complexity of the brute force solution is `O(n^2)`. To improve efficiency, you can sort the array, and then use **two pointers**, one pointing to the head of the array and the other pointing to the tail of the array, and decide `left += 1` or `right -= 1` according to the comparison of `sum` and `target`.
2. After finding the two values which `sum` is `target`, you can use the `index()` method to find the `index` corresponding to the value.

## Complexity

- Time complexity: `O(N * log N)`.
- Space complexity: `O(N)`.

## Python

```python
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        original_nums = nums.copy()
        nums.sort()

        left = 0
        right = len(nums) - 1

        while left < right:
            sum_ = nums[left] + nums[right]

            if sum_ == target:
                break

            if sum_ < target:
                left += 1
                continue

            right -= 1

        return [
            original_nums.index(nums[left]),
            len(nums) - 1 - original_nums[::-1].index(nums[right])
        ]
```

## Other languages

```java
// Welcome to create a PR to complete the code of this language, thanks!
```

Dear LeetCoders! For a better LeetCode problem-solving experience, please visit website [leetcoder.net](https://leetcoder.net): Dare to claim the best practices of LeetCode solutions! Will save you a lot of time!

Original link: [1. Two Sum - LeetCode Best Practices](https://leetcoder.net/en/leetcode/1-two-sum).

GitHub repository: [f*ck-leetcode](https://github.com/fuck-leetcode/fuck-leetcode).