-[1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance](en/1001-2000/1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.md) was solved in_Python_.

-[433. Minimum Genetic Mutation](en/1-1000/433-minimum-genetic-mutation.md) was solved in_Python_ and2 ways.

-[433. Minimum Genetic Mutation](en/1-1000/433-minimum-genetic-mutation.md) was solved in_Python_ and3 ways.

TODO: Add A* for433 and127?

TODO: Add A* for 127?

More LeetCode problems will be added soon.

LeetCode link:[454. 4Sum II](https://leetcode.com/problems/4sum-ii),

[454. 四数相加 II](https://leetcode.cn/problems/4sum-ii)

LeetCode link:[454. 4Sum II](https://leetcode.com/problems/problem), difficulty:**Medium**.

[中文题解](#中文题解)

Given four integer arrays`nums1`,`nums2`,`nums3`, and`nums4` all of length`n`, return the number of tuples` (i, j, k, l)` such that:

Difficulty:**Medium**

**Input**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

[中文题解](#中文题解)

1. Because the final requirement is to take one number from each group of numbers, the four groups of numbers can be split into**two groups of two**.

2. Count the number of each`sum`. Use`Map` to store,`key` is`sum`,`value` is`count`.

3. Iterate over`nums3` and`nums4`, if`-(num3 + num4)` exists in`keys` of`Map`, then`count` is included in the total.

给你四个整数数组`nums1`、`nums2`、`nums3` 和`nums4` ，数组长度都是`n` ，请你计算有多少个元组`(i, j, k, l)` 能满足：

难度:**中等**

**输入**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

**输出**:`2`

**解释**:

1. 因为最终要求是每组数中各取一个数，所以可以把四组数拆分成两个两组数。

2. 统计出每个`和`值对应的个数。使用`Map`储存，`key`为`和`，`value`为`count`。

3. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

1. 统计出每个`和`值对应的个数。使用`Map`储存，`key`为`和`，`value`为`count`。

num_to_count= defaultdict(int)

for num1in nums1:

for num2in nums2:

num_to_count[num1+ num2]+=1

2. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

result=0

for num3in nums3:

for num4in nums4:

result+= num_to_count[-(num3+ num4)]

return result

LeetCode link:[454. 4Sum II](https://leetcode.com/problems/4sum-ii),

[454. 四数相加 II](https://leetcode.cn/problems/4sum-ii)

力扣链接：[454. 四数相加 II](https://leetcode.cn/problems/4sum-ii) ，难度：**中等**。

[中文题解](#中文题解)

Given four integer arrays`nums1`,`nums2`,`nums3`, and`nums4` all of length`n`, return the number of tuples` (i, j, k, l)` such that:

给你四个整数数组`nums1`、`nums2`、`nums3` 和`nums4` ，数组长度都是`n` ，请你计算有多少个元组`(i, j, k, l)` 能满足：

Difficulty:**Medium**

**Input**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

**输入**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

**Output**:`2`

**输出**:`2`

**Explanation**:

**解释**:

**Input**:`nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]`

**输入**:`nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]`

**Output**:`1`

**输出**:`1`

[中文题解](#中文题解)

1. Because the final requirement is to take one number from each group of numbers, the four groups of numbers can be split into**two groups of two**.

2. Count the number of each`sum`. Use`Map` to store,`key` is`sum`,`value` is`count`.

3. Iterate over`nums3` and`nums4`, if`-(num3 + num4)` exists in`keys` of`Map`, then`count` is included in the total.

1. 因为最终要求是每组数中各取一个数，所以可以把四组数拆分成两个两组数。

2. 统计出每个`和`值对应的个数。使用`Map`储存，`key`为`和`，`value`为`count`。

3. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

num_to_count= defaultdict(int)

num_to_count[num1+ num2]+=1

result=0

return result

*Time:`O(n * n)`.

*Space:`O(n)`.

*时间:`O(n * n)`.

*空间:`O(n)`.

给你四个整数数组`nums1`、`nums2`、`nums3` 和`nums4` ，数组长度都是`n` ，请你计算有多少个元组`(i, j, k, l)` 能满足：

难度:**中等**

**输入**:`nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]`

**输出**:`2`

**解释**:

1. 因为最终要求是每组数中各取一个数，所以可以把四组数拆分成两个两组数。

2. 统计出每个`和`值对应的个数。使用`Map`储存，`key`为`和`，`value`为`count`。

3. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

1. 统计出每个`和`值对应的个数。使用`Map`储存，`key`为`和`，`value`为`count`。

num_to_count= defaultdict(int)

for num1in nums1:

for num2in nums2:

num_to_count[num1+ num2]+=1

2. 遍历`nums3`和`nums4`，如果`-(num3 + num4)`存在于`Map`的`keys`中，则`count`计入总数。

result=0

for num3in nums3:

for num4in nums4:

result+= num_to_count[-(num3+ num4)]

return result