Category: Python Coding Test

In computer science, algorithm problems are important elements that provide us with efficient methods to solve problems.
Among them, the sliding window technique is used to find continuous subarrays that satisfy certain conditions in linear data structures like arrays or strings.
In this lecture, we will solve a problem using the sliding window technique and examine how this technique is useful for problem-solving.

Problem Description

Given an integer array nums and an integer k, calculate the sum of the subarray with the maximum sum among subarrays of size k.
In other words, this is a problem of maximizing the sum of k consecutive elements.
For example, if the array is nums = [1, 3, 2, 5, 4, 8] and k = 3,
the sum of the subarray [5, 4, 8] is the largest, so the answer is 17.

Problem Requirements

• Function input: integer array nums and integer k
• Function output: maximum sum of the subarray

Approach to the Problem

By using the sliding window technique to solve this problem, we can reduce the time complexity to O(n) by traversing the input array only once.
The idea of this technique is to use two pointers to adjust the start and end of the current window to calculate the sum of the subarray.

Explanation of the Sliding Window Technique

1. Define the initial window. Use pointers to select the first k elements and calculate their sum.
2. Then, move to the second window by removing one element from the start of the window and adding one element to the end.
   Repeat this process until the end of the array.
3. Record the sum obtained at each step and update the maximum sum by comparing it with the previously recorded maximum sum.

Code Implementation

def max_sum_subarray(nums, k):
    # Calculate initial window
    max_sum = sum(nums[:k])
    window_sum = max_sum
    
    # Apply sliding window
    for i in range(k, len(nums)):
        # Remove the leftmost number from the current window and add the newly added number.
        window_sum += nums[i] - nums[i - k]
        max_sum = max(max_sum, window_sum)
    
    return max_sum

# Example execution
nums = [1, 3, 2, 5, 4, 8]
k = 3
result = max_sum_subarray(nums, k)
print(f"Maximum sum of the subarray: {result}")

Code Explanation

The above function max_sum_subarray takes an array nums and an integer k as arguments and returns the maximum sum.
First, it calculates the sum of the initial window and then traverses the array using the sliding window method.
The sum of each window is obtained by removing the leftmost element from the previous sum and adding a new element,
recording each window’s sum to update the maximum sum.

Results and Testing

When you run the above example, the result Maximum sum of the subarray: 17 is produced.
By utilizing the sliding window technique, we can solve the problem quickly with just one traversal.

Conclusion

In this lecture, we solved the problem of finding the maximum subarray sum using the sliding window technique.
This technique is very useful as it reduces time complexity by traversing the entire array only once without the need to repeatedly compare the same elements.
It can also be applied to various other problems, making it a great help in areas where coding tests and algorithm problems are frequently presented.

Table of Contents

1. 1. Introduction
2. 2. Stack
3. 3. Queue
4. 4. Problem Description
5. 5. Solution Process
6. 6. Conclusion

1. Introduction

Two of the most fundamental data structures in learning programming are stacks and queues. These data structures have their own consistent operating principles and are frequently used in various algorithm problems. In this article, we will explain the concepts of stacks and queues and discuss algorithm problems that can be solved using them.

2. Stack

A stack is a data structure with a Last In First Out (LIFO) structure. The most recently inserted data is the first to be removed. The main operations of a stack include:

• push(item): Adds item to the top of the stack.
• pop(): Removes and returns the item at the top of the stack.
• peek(): Returns the item at the top of the stack without removing it.
• is_empty(): Returns True if the stack is empty, otherwise returns False.

3. Queue

A queue is a data structure with a First In First Out (FIFO) structure. The first data inserted is the first to be removed. The main operations of a queue are as follows:

• enqueue(item): Adds item to the end of the queue.
• dequeue(): Removes and returns the item at the front of the queue.
• peek(): Returns the item at the front of the queue without removing it.
• is_empty(): Returns True if the queue is empty, otherwise returns False.

4. Problem Description

The problem we will solve this time is Validating Parentheses using Stacks and Queues. The problem is as follows:

Write a function to check if a given string consisting only of parentheses is valid. A valid parentheses expression is one where all opened parentheses are closed correctly and each closing parenthesis matches an opened parenthesis.

Input

The input will be a single string consisting of the following parentheses:

• ‘(‘
• ‘)’
• ‘{‘
• ‘}’
• ‘[‘
• ‘]’

Output

Return True if all parentheses are correctly closed, otherwise return False.

5. Solution Process

To solve this problem, we will use a stack. We will process each character of the input string, adding opened parentheses to the stack and checking for matching opened parentheses when closing parentheses appear. The following is the process of solving the problem:

Step 1: Initialize the Stack

We initialize an empty list to use as a stack to traverse each character of the string.

Step 2: Set Up Parentheses Mapping

We create a dictionary to define the relationship between opened and closed parentheses.

Step 3: Traverse the String

We traverse the string character by character, adding opened parentheses to the stack. If a closing parenthesis appears, we check for matching with the top element of the stack.

Step 4: Check Stack Status

After traversing the string, if there are no contents left in the stack, it indicates that all parentheses are correctly closed.

Python Code Implementation

def is_valid_parentheses(s: str) -> bool:
    stack = []
    mapping = {")": "(", "}": "{", "]": "["}
    
    for char in s:
        if char in mapping:  # In case of a closing parenthesis
            top_element = stack.pop() if stack else '#'  # Pop from the stack
            if mapping[char] != top_element:  # Check mapping
                return False
        else:  # In case of an opening parenthesis
            stack.append(char)  # Add to the stack
    
    return not stack  # If the stack is empty, return True, otherwise return False
    

Step-by-Step Explanation

Now, let’s delve into each step in more detail.

Stack Initialization

We initialize the stack as an empty list with “stack = []“. The stack is used to check if the parentheses are opened and closed in the correct order.

Setting Up Parentheses Mapping

We define the mapping of each closing parenthesis to its corresponding opening parenthesis by creating a dictionary: “mapping = {“)”: “(“, “}”: “{“, “]”: “[“}“. This mapping is used for comparison when encountering closing parentheses.

Traversing the String

We use a for loop to traverse the string character by character. If the character is a closing parenthesis, we pop an element from the stack and compare it with the mapped opening parenthesis. If there is no match, it is invalid, and we return False. If the character is an opening parenthesis, we add it to the stack.

Checking Stack Status

If the stack is empty after complete traversal, it indicates that all opened parentheses are matched correctly. Thus, return not stack returns True if the stack is empty, otherwise it returns False.

6. Conclusion

In this tutorial, we first examined the basic concepts of stacks and queues, and then discussed the process of solving the parentheses validation problem using a stack. Stacks, with their LIFO structure, are useful in various algorithm problems, and understanding and utilizing them is very important for preparing for coding tests.

Queues are also important data structures that operate in a FIFO manner and can be applied to many problems. Stacks and queues form the basis for solving coding problems, and a good understanding of these data structures is often required. Based on the content discussed in this article, I hope you practice by solving more problems to build your skills.

Creation Date: October 10, 2023

Author: Algorithm Instructor

Problem Description

Given an integer N, create a sequence of numbers sorted in ascending order from 1 to N using a stack. You are allowed to perform the following operations:

• Push the numbers from 1 to N onto the stack in order.
• Pop the value on the top of the stack and print it.

You will receive two numbers as input:

• Integer N (1 <= N <= 100,000)
• Integer K (1 <= K <= N)

The output should be K numbers printed in ascending order. You can use the stack, and you must appropriately push and pop the given input numbers from the stack. Your goal is to generate a sequence in ascending order using the stack operations.

Problem Approach

This problem utilizes the Last-In-First-Out (LIFO) characteristic of stacks to output the given numbers in ascending order. First, you should push the numbers from 1 to N onto the stack, then pop the required numbers in order to output them. There are several important points to consider in this process:

• The numbers that can be pushed to the stack are from 1 to N.
• When popping from the stack, you must always pop from the topmost number.
• The final output numbers must be sorted in ascending order.

Algorithm Implementation

Now, let’s implement the algorithm based on the problem approach. The basic logic of the algorithm is as follows:

1. Push the numbers up to N onto the stack in order.
2. Pop the top number from the stack and output the desired K numbers.

Below is the code implemented in Python:

def create_sorted_sequence(N, K):
    stack = []
    result = []
    num = 1
    
    for i in range(N):
        while num <= N:
            stack.append(num)
            num += 1
        
        if stack:  # If the stack is not empty
            result.append(stack.pop())
        
        if len(result) == K:  # Stop when K numbers have been output
            break
            
    return result

# Example input
N = 5
K = 3
output = create_sorted_sequence(N, K)
print("Ascending sequence:", output)

            

The function create_sorted_sequence(N, K) takes N and K as inputs and generates a sequence of K numbers in ascending order. It stores the numbers using the stack and pops them to add to the result array as needed.

Code Explanation

Let’s explain each part of the code in detail:

1. Initializing the stack and result array:

   stack = [] and result = [] are used to initialize an empty stack and result array.

2. Number Push Logic:

   The condition while num <= N: is used to push numbers from 1 to N onto the stack. The num variable determines the next number to be pushed.

3. Number Pop Logic:

   The if stack: is used to pop the top number from the stack and add it to the result array if the stack is not empty.

4. K Output Condition:

   The condition if len(result) == K: checks if the number of output amounts has reached K, and if so, the loop is exited.

Time Complexity

The time complexity of this algorithm is as follows:

• Process of pushing numbers onto the stack: O(N)
• Process of popping K numbers from the stack: O(K)

Thus, the overall time complexity is O(N + K). This is an efficient approach and performs quickly enough.

Conclusion

The problem of sorting numbers in ascending order using a stack requires a creative and systematic approach. Through this lecture, you learned the basic principles of stacks and how to solve problems using them. I hope you continue to improve your skills by tackling various algorithm problems.

Hello! In this coding test lecture, we will tackle the problem of finding the sum of numbers using Python. This problem is a basic type that frequently appears in algorithm problem-solving, and it is useful for practicing basic input handling and calculations using loops.

Problem Description

Write a function sum_numbers(N) that calculates the sum of all integers from 1 to N for a given integer N. Additionally, you should prompt the user to input the value of N and print the corresponding sum.

Input

• Integer N (1 ≤ N ≤ 10000)

Output

• The sum of all integers from 1 to N

Example

Input Example

5

Output Example

15

Problem Solving Process

To solve this problem, we will go through the following steps.

1. Understanding the Problem

Let’s organize the approach to understand the problem. We need to compute the sum of integers from 1 to N, which can be done using loops or a mathematical formula. Using a loop allows us to add each number iteratively, while using a mathematical formula can yield a more efficient solution.

2. Mathematical Approach

The sum of integers from 1 to N can be calculated mathematically using the following formula:

Sum = N * (N + 1) / 2

This formula allows us to obtain the result with O(1) time complexity instead of summing all integers.

3. Implementing Python Code

Now, let’s implement this problem in Python. We will implement it in two ways: using a loop and using a mathematical formula.

Method 1: Using a Loop

def sum_numbers_loop(N):
    total = 0
    for i in range(1, N + 1):
        total += i
    return total

# Get user input
N = int(input("Please enter an integer (1 ≤ N ≤ 10000): "))
result = sum_numbers_loop(N)
print("The sum from 1 to", N, "is:", result)

Method 2: Using a Mathematical Formula

def sum_numbers_formula(N):
    return N * (N + 1) // 2

# Get user input
N = int(input("Please enter an integer (1 ≤ N ≤ 10000): "))
result = sum_numbers_formula(N)
print("The sum from 1 to", N, "is:", result)

4. Testing and Exception Handling

To verify that our functions work correctly, we will create several test cases. For example, we can test the boundary values of N (1 and 10000) and some general values.

Test Cases

• Input: 1, Output: 1
• Input: 5, Output: 15 (1+2+3+4+5)
• Input: 10000, Output: 50005000 (1+2+…+10000)

In addition to these, we can ensure the stability of the function through a variety of inputs.

5. Complexity Analysis

The time complexity for each method is as follows:

• Using a loop: O(N) – Performance may degrade as N increases.
• Using a mathematical formula: O(1) – Much faster and more efficient.

Conclusion

In this lecture, we explored two methods to solve the problem of finding the sum of numbers in Python. While we can solve the problem using a loop, we learned that utilizing a mathematical formula is far more efficient. Through these basic problems, we can build our coding foundations and prepare to tackle more complex challenges.

Additional Learning Resources

• Baekjoon Online Judge – A platform to solve various algorithm problems
• Programmers – Algorithm problems and solutions for interview preparation
• Codewars – A website where you can build your skills by solving problems

Thank you! We hope you continue to build your skills through a variety of algorithm problems. See you in the next lecture!

Problem Description

Given a number N, we need to generate all permutations of numbers from 1 to N and find the order of a specific permutation. For example, when N=3, the possible permutations are [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1], totaling 6, and we need to find the position of each permutation starting from index 1.

To solve this problem, we will receive the following input:

• N: The number of elements to form the permutation
• P: A specific permutation (provided in list format)

Here, the output will be the order of the given permutation P.

Example Input and Output

Input

3
[2, 3, 1]
            

Output

4
            

In the above case, the given permutation [2, 3, 1] is the 4th permutation out of a total of 6 permutations.

Problem-Solving Process

1. Generating Permutations

To solve the problem, we will first use the permutations function from Python’s itertools module to generate all permutations from 1 to N. The permutations function is very useful for returning all permutations of a given iterable.

2. Sorting into a List

The generated permutations will be stored in a list to maintain a sorted form. This is to quickly find permutations indexed from 1.

3. Finding the Index of a Specific Permutation

We will search for the given specific permutation in the list to find its index. Since the index starts from 0, we will add 1 when outputting the result.

Python Code Implementation

Now, let’s implement the Python code based on the approach described above:

from itertools import permutations

def find_permutation_index(N, P):
    # Create a list of numbers from 1 to N
    numbers = list(range(1, N+1))
    
    # Generate all permutations
    all_permutations = list(permutations(numbers))

    # Find the index of the specific permutation
    index = all_permutations.index(tuple(P)) + 1  # Add 1 to adjust to 1-based indexing
    return index

# Example execution
N = 3
P = [2, 3, 1]
print(find_permutation_index(N, P))
            

The above code defines the find_permutation_index function, which finds the index of a specific permutation given N and P. We utilize the itertools module to automatically generate all permutations and easily find the position of the permutation using the index method.

Complexity Analysis

The time complexity of this algorithm is proportional to the number of permutations generated, which is N!. While this may be an inefficient approach, it is easy to understand and implement when N is small. However, an efficient approach is needed for larger values of N.

Improved Approach

For example, we could use a mathematical approach to collect counts for each number and systematically calculate how many combinations each specific number can generate. This could be a more efficient method.

Conclusion

In this tutorial, we learned how to solve the problem of finding the order of permutations. We presented a simple implementation using Python’s itertools module, so apply it to real problems for a deeper understanding.