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

Original link:[leetcoder.net - LeetCoder: Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/1-two-sum)

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

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.

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

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

**Explanation**：`Because nums[0] + nums[1] == 9, we return [0, 1].`

**Explanation**:Because nums[0] + nums[1] == 9, we return[0, 1].

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

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

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

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

-**Only one valid answer exists.**

<details>

<summary>Hint 1</summary>

Hint 1

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>

Hint 2

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>

Hint 3

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>

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:

- Solution 1:Use`index()` method to find it.

- Solution 2:Bring the`index` when sorting, that is, the objecttobe sorted is an array of tuples of`(num, index)`.

- 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()` methodtofind it. I have discussed this in another solution.

* Time:`O(N * log N)`.

* Space:`O(N)`.

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

- Space complexity:`O(N)`.

classSolution:

deftwoSum(self,nums: List[int],target:int) -> List[int]:

num_index_list= [(num, i)for i, numinenumerate(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

right-=1

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`.

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

let numToIndex=newMap()

for (let i=0; i<nums.length; i++) {

numToIndex.set(nums[i], i)

}

```

let numToIndex=newMap()

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`.

let numToIndex=newMap()

if (numToIndex.has(target-nums[i])) {// 1

return [numToIndex.get(target- nums[i]), i]//2

### Complexity

* Time:`O(n)`.

* Space:`O(n)`.

- Time complexity:`O(N)`.

- Space complexity:`O(N)`.

### Solution1: Two pointers

### Solution2: UsingMap

classSolution {

publicint[]twoSum(int[]nums,inttarget) {

var numToIndex=newHashMap<Integer,Integer>();

for (var i=0; i< nums.length; i++) {

if (numToIndex.containsKey(target- nums[i])) {

returnnewint[]{numToIndex.get(target- nums[i]), i};

### Solution1: Two pointers

### Solution2: UsingMap

classSolution:

deftwoSum(self,nums: List[int],target:int) -> List[int]:

num_to_index= {}

for i, numinenumerate(nums):

if target- numin num_to_index:

return [num_to_index[target- num], i]

### Solution1: Two pointers

### Solution2: UsingMap

classSolution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

unordered_map<int, int> num_to_index;

### Solution1: Two pointers

### Solution2: UsingMap

vartwoSum=function (nums,target) {

let numToIndex=newMap()

for (let i=0; i<nums.length; i++) {

if (numToIndex.has(target- nums[i])) {

return [numToIndex.get(target- nums[i]), i]

### Solution1: Two pointers

### Solution2: UsingMap

publicclassSolution {

publicint[]TwoSum(int[]nums,inttarget) {

varnumToIndex=newDictionary<int,int>();

for (inti=0;i<nums.Length;i++) {

if (numToIndex.ContainsKey(target-nums[i])) {

return [numToIndex[target-nums[i]],i];

### Solution1: Two pointers

### Solution2: UsingMap

functwoSum(nums []int,targetint) []int {

numToIndex:=map[int]int{}

### Solution1: Two pointers

### Solution2: UsingMap

deftwo_sum(nums,target)

num_to_index= {}

##C, Kotlin, Swift, Rust or other languages

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.

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

- Space complexity:`O(N)`.

classSolution:

deftwoSum(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:

if sum_< target:

left+=1

right-=1

return [

original_nums.index(nums[left]),

len(nums)-1- original_nums[::-1].index(nums[right])

]