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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

By the time we traverse the complete list (for creating the hash table), we can find the list length. I ct us say the list length is $M$. To find $n^{\text {th }}$ from the end of linked list, we can convent this to $(M-n+1)^{\text {th }}$ from the beginuing. Sance we already know the length of the list, it is just a matter of retuming $(M-n+1)^{\text {th }}$ key value from the hash table.
‘ Iine Complexity: I inve lor creating the hash table, $T(m)=()(m)$. Space Cionaplexity: Since we need to create a hash table of size $m$, O( $m)$.
Problem-4 Can we use Problem-3 approach for solving Problem-2 without creating the hash table?
Solution Yeg. If we observe the Problenh3 solution, what we are actually doing is finding the size of the linked list. That means we are using the hash table to find the size of the linked list. We can find the length of the linked list just by starting at the head node and traversing the list. So, we can find the length of the list without creating the hash table. After finding the length, compute $M-n+1$ and with one more scan we can get the $(M-n+1)^{\text {th }}$ node from the beginning. This solution needs two scans: one for finding the length of the list and the other for finding $(M-n+1)^{t h}$ node from the hegianing-
Time Complexity: Time for finding the length + Time for finding the $(M-n+1)^{\text {th }}$ node from the beginning. Therefore, $T(n=O(n)+$ $O(n) \approx O(n)$. Space Complexity: $O(1)$. Hence, no need to create the hash table.
Problem5 Can we solve Problen-2 in one scan?
Solution: Yes. Efficient Approach: Use two pointers $p N$ thNode and $p$ Temp. Initially, both point to head node of the list. $p N$ thNode starts moving only alter $p$ Temp has made $n$ mones. From there both mowe forward until $p$ Temp reaches the end of the list. As a result, $p N$ thNode points to $n^{\text {th }}$ node from the end of the linked list.
Notet At any poant of time both move one node at a time.

## 统计代写|数据结构作业代写data structure代考|Efficient Approach

Solution: Yes. Fincient Approed QMemoryleas Appronch): The space conplexity can be reduced to O(1) by considering two pointers at differeat speed – a slow pointer and a fast pointer. The slow poanter moves one step at a time while the fast pointer mowes two steps at a tine. This problem was solved by Floyd. The solution is named the Floyd cycle finding algorithm. It uses two pointers moving at different speeds to walk the linked list. If there is no cycle in the list, the fast poanter will eventually reach the ead and we can return false in this case. Now consider a cyclic list and imagiae the slow and fast pointers are two rumers racing around a circle track. Once they enter the loop they are expected to neet, which denotes that there is a loop.

This works becanse the only way a faster movang pointer would point to the sanae location as a slower moving pointer is if sonachow the entire list or a part of it is circular. Think of a tortoise and a hare rumaing on a track. The faster numang hare will catch up with the tortoise if they are ruming in a loop. As an exanple, consider the followanng exmple and trace out the Floyd algonthm. From the diagrans below we can see that after the final step they are meeting at sone point in the loop which nay not be the starting point of the loop.
Note: slowPtr (tortoise) moves one pointer at a tine and fastPtr (hare) mones two pointers at a tine.

## 统计代写|数据结构作业代写data structure代考|Algorithm

Create two stacks one for the first list and one for the second list.
Traverse the first list and push all the node addresses onto the first stack.
Traverse the second list and push all the node addresses onto the second stack.
Now both stacks contain the node address of the corresponding lists.
Now compare the top node address of both stacks.
If they are the same, take the top elements from both the slacks and keep them in some temporary variable (since both node addresses are node, it is enough if we use one temporary varable).
Continue this process until the top node addresses of the stacks are not the same.
This point is the one where the lists merge into a single list.
Return the value of the tenporary variable.

## 数据结构代写

‘ 线复杂性：我想创建哈希表，吨(米)=()(米). Space Cionaplexity：因为我们需要创建一个大小为米， 这（米).

## 广义线性模型代考

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。