Author: root

1. Problem Definition

The interval sum problem, which is frequently asked in coding tests, is about efficiently calculating the sum of elements in a specific range within a given array. In this post, we will discuss the ‘Calculating Interval Sum 1’ problem.

Problem Description

An integer N and M are given, along with an array consisting of N integers. Next, M queries are provided, each consisting of two integers S and E. The task is to calculate the sum from S to E. Note that S and E range from 1 to N.

2. Input and Output Format

Input

• The first line contains N and M separated by a space. (1 ≤ N, M ≤ 100,000)
• The second line contains N integers separated by spaces. (Each integer is between 1 and 100,000)
• From the third line onward, M lines contain two queries. (S, E)

Output

• Output the answer for each query over M lines.

3. Sample Input and Output

Sample Input

    5 3
    1 2 3 4 5
    1 3
    2 4
    3 5
    

Sample Output

    6
    9
    12
    

4. Approach to the Problem

To calculate the interval sum, the simplest method can be used first. However, since the maximum values of N and M are 100,000, it is impossible with a time complexity of O(N * M). Therefore, we need to find an efficient way to calculate the interval sum.

4.1. Simple Approach

The simplest method is to iterate from S to E to calculate the sum. However, this method has a complexity of O(N * M).

4.2. Utilizing Cumulative Sum Array

One of the efficient approaches is to use a Cumulative Sum array. By creating the cumulative sum array first, we quickly calculate the answer for each query using the values of S and E.

When defining the cumulative sum array, the i-th element of the array represents the sum from 1 to i. That is, the i-th value of the cumulative sum array is calculated as:

    sum[i] = sum[i - 1] + array[i - 1]
    

When processing queries, the following formula can be used:

    result = sum[E] - sum[S - 1]
    

5. Algorithm Implementation

Now let’s implement the algorithm using the cumulative sum array described above.

5.1. Python Code

def solve(N, M, array, queries):
    # Create cumulative sum array
    sum_array = [0] * (N + 1)
    
    # Calculate cumulative sum
    for i in range(1, N + 1):
        sum_array[i] = sum_array[i - 1] + array[i - 1]
    
    result = []
    
    # Process queries
    for S, E in queries:
        query_sum = sum_array[E] - sum_array[S - 1]
        result.append(query_sum)
    
    return result

# Sample input
N, M = 5, 3
array = [1, 2, 3, 4, 5]
queries = [(1, 3), (2, 4), (3, 5)]

# Output results
output = solve(N, M, array, queries)
for res in output:
    print(res)
    

6. Code Explanation

The above code works in the following order:

1. First, the cumulative sum array is initialized. The size of the cumulative sum array is set to N + 1 to easily calculate the sum from 1 to N.
2. Next, each element of the array is iterated to calculate the cumulative sum.
3. While going through the queries, the sum for the given S and E is quickly calculated using the cumulative sum array.
4. The calculated results are stored in a list to be output at the end.

7. Time Complexity Analysis

The time complexity of the above algorithm is as follows:

• O(N) to create the cumulative sum array
• O(M) to process M queries

Thus, the overall time complexity is O(N + M), allowing us to solve the problem efficiently.

8. Space Complexity Analysis

The space complexity is O(N) to store the cumulative sum array. Additional variables use constant space, so the overall space complexity is O(N).

9. Conclusion

In this post, we discussed the interval sum problem. We learned how to solve the problem efficiently using a cumulative sum array and provided a Python code example to aid understanding. This method can be applied in various problems requiring interval sums and can serve as a basis to tackle more complex issues.

10. Additional Exercises

Now, try to solve some exercises on your own. Below are a few additional practice problems.

• Write a program to calculate the interval sum from 1 to 100.
• Implement a program that calculates the sum of the interval each time an element changes.
• Write a program to calculate the sum of a specific area in a two-dimensional array.

11. Questions and Feedback

If you have any questions or feedback regarding the content discussed in this article, feel free to leave a comment. Let’s solve problems together!

Hello, everyone! Today we will tackle the range sum problem, which frequently appears in coding tests using Python. The range sum problem asks how to quickly calculate the sum of a specific range, and it’s an important concept that forms the basis of many algorithmic problems. We will explore various approaches to solve this problem.

Problem Description

Given an array as follows, write a program to efficiently calculate the sum of specific ranges for multiple query pairs.

Array: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Query: [(1, 3), (2, 5), (3, 7)]

When the above array and queries are provided, the queries are given in the form of (start index, end index). For example, the query (1, 3) means that we need to calculate the sum from index 1 to index 3 of the array (1-based index). In this case, it should be 2 + 3 + 4 = 9.

Approach to the Problem

There are various basic approaches to solve the range sum problem. I will start with the simplest method and explain more efficient ones.

1. Using Simple Iteration

The most intuitive method is to calculate the sum for the required range directly. This method is good when there are few queries, but it is inefficient when there are many. We can implement it as follows.

def simple_range_sum(arr, queries):
    results = []
    for start, end in queries:
        results.append(sum(arr[start - 1:end]))  # 1-based index to 0-based index
    return results


arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
queries = [(1, 3), (2, 5), (3, 7)]
print(simple_range_sum(arr, queries))  # Output: [9, 14, 27]

When we execute the code above, the results for the queries will output [9, 14, 27]. However, the time complexity of this method is O(Q * N), where Q is the number of queries and N is the size of the array. This becomes inefficient for large inputs.

2. Using a Cumulative Sum Array

A more efficient way is to create a cumulative sum array. Once we have the cumulative sum array, we can calculate the sum of each range in O(1) time. The method is as follows.

def prefix_sum(arr):
    n = len(arr)
    prefix = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix[i] = prefix[i - 1] + arr[i - 1]
    return prefix


def efficient_range_sum(arr, queries):
    prefix = prefix_sum(arr)
    results = []
    for start, end in queries:
        results.append(prefix[end] - prefix[start - 1])  # 1-based index adjustment
    return results


arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
queries = [(1, 3), (2, 5), (3, 7)]
print(efficient_range_sum(arr, queries))  # Output: [9, 14, 27]

In the above code, calculating the cumulative sum array takes O(N) time, and processing each query takes O(1) time. Thus, the overall time complexity reduces to O(N + Q). This method allows us to handle multiple queries efficiently.

Problem Optimization and Selection

The range sum problem can be utilized in various situations, and the algorithm chosen may vary depending on the conditions of the problem. It is essential to choose the optimal method considering the size of the input array, the number of queries, and the length of each query’s range.

Besides the methods described above, more complex data structures such as segment trees can also be used, but this tutorial focused on basic methods for solving practical problems. In the next lesson, we will address more complicated dynamic query problems.

Conclusion

In this lesson, we covered the range sum problem using Python. We found that while simple iteration can be inefficient, using a cumulative sum array allows us to solve the problem much more efficiently. The range sum problem is a very useful topic for building basic algorithm skills, so I encourage you to practice it multiple times and try various variations!

Python Coding Test Course: Calculating the Product of Intervals

When solving programming problems, there are often cases where calculations are performed based on specific ranges of given values. In this article, we will address the problem of ‘calculating the product of intervals.’ This problem involves implementing an algorithm that multiplies the values belonging to a specific range within a given array. During this process, we will also discuss strategies to solve the problem efficiently.

Problem Description

There is a given array of integers and several queries. Each query consists of two indices (l, r), and the task is to output the product of all elements in the array within that range, including the specified indices. It is important to note that the result of the product of the interval can be a very large number, so it needs to be optimized for calculation.

Example:
Array: [1, 2, 3, 4, 5]
Query: (1, 3)
Output: 2 * 3 * 4 = 24

Problem-Solving Strategy

1. Basic Approach

The most intuitive approach is to iterate over the elements at the specified indices for each query in the given array and multiply them. However, this method has a worst-case time complexity of O(Q * N), which is inefficient. Here, N represents the size of the array and Q is the number of queries. This method can become significantly slower as the number of queries increases.

2. Approach Using Cumulative Product

One efficient method is to use the cumulative product (Prefix Product). The way to calculate the cumulative product is to compute and store the product of all previous elements for each element in the array.

• For example, if the array is A = [1, 2, 3, 4, 5]:

  Cumulative product array P:
  P[0] = 1
  P[1] = 1 * 1 = 1
  P[2] = 1 * 1 * 2 = 2
  P[3] = 1 * 1 * 2 * 3 = 6
  P[4] = 1 * 1 * 2 * 3 * 4 = 24
  P[5] = 1 * 1 * 2 * 3 * 4 * 5 = 120
  

After constructing the cumulative product array, the product of the interval can be calculated using the following formula:

Interval Product = P[r+1] / P[l]

Here, P[i] represents the product up to the i-th element. By using this method, we can calculate the product of the interval with a time complexity of O(1).

Python Code Implementation

Now, let’s implement the algorithm for calculating the product of intervals in Python, referring to the approach above.

def calculate_prefix_product(arr):
    n = len(arr)
    prefix_product = [1] * (n + 1)
    
    for i in range(n):
        prefix_product[i + 1] = prefix_product[i] * arr[i]
    
    return prefix_product

def range_product(arr, queries):
    prefix_product = calculate_prefix_product(arr)
    results = []
    
    for l, r in queries:
        product = prefix_product[r + 1] // prefix_product[l]
        results.append(product)
    
    return results

# Example array and queries
arr = [1, 2, 3, 4, 5]
queries = [(1, 3), (0, 2), (2, 4)]

# Function call
result = range_product(arr, queries)
print(result)  # Output: [24, 6, 60]

Final Check

Now it’s time to verify if the code works properly. We will check if the correct results are obtained through various test cases. It is important to ensure that the range of queries is valid and that there are no issues at boundary values.

• Test Case 1: arr = [1,2,3,4,5], queries = [(0, 4)] => Result: 120
• Test Case 2: arr = [10,20,30], queries = [(0, 2), (1, 1)] => Result: [6000, 20]
• Test Case 3: arr = [0, 1, 2, 3], queries = [(0, 3), (1, 1)] => Result: [0, 1]

Conclusion

The problem of calculating the product of intervals discussed in this tutorial demonstrates how to derive an efficient solution by appropriately utilizing cumulative products. This principle can also be applied to solve other similar problems. Practicing various algorithmic problems will enhance your preparation for coding tests. Make good use of this technique in your future coding tests!

Additional Questions

If you have any unresolved issues or additional questions, feel free to reach out to me at any time. Happy Coding!

Hello! Today we will take a closer look at one of the algorithm problems that often appears in Python coding tests, “Counting Stair Numbers”. Through this problem, we will learn the concept of dynamic programming and the process of solving problems.

Problem Description

A stair number is defined by the following rules:

• A stair number is composed of digits from 0 to 9.
• The digits of two adjacent positions must differ by exactly 1. For example, 234 is valid, but 235 or 123 are not valid.
• The first digit of a stair number cannot be 0.

Given a natural number N, how many stair numbers of length N exist? For example, if N is 2, there are 9 possible stair numbers: 10, 12, 21, 23, 32, 34, 43, 45, 54, 65, 76, 78, 87, 89, 98, totaling 15.

Input Format

The first line contains the number of digits in the stair number N (1 ≤ N ≤ 100).

Output Format

The first line should output the number of N-digit stair numbers modulo 1,000,000,000.

Example Input

2

Example Output

15

Problem Solving Strategy

This problem can be solved using dynamic programming. Stair numbers can be categorized according to the rules of the problem, and for each digit, we consider the possible number combinations to derive the results. Now, let’s look at the specific solution process.

Step 1: State Definition

To find N-digit stair numbers, we define a DP array. We use dp[n][k] to represent the number of stair numbers of length n whose last digit is k. Here, n is the number of digits, and k is the last digit (0 to 9).

Step 2: Initial Condition Setting

The stair numbers of length 1 are from 1 to 9. Therefore, we set dp[1][0] = 0 (0 cannot be the first digit), and dp[1][1] = dp[1][2] = ... = dp[1][9] = 1.

Step 3: Deriving the Recursion Formula

To construct stair numbers of length n, we add one digit to stair numbers of length n-1. If the last digit is 0, it can lead to 1, and if the last digit is 9, it can lead to 8. Therefore, we get the following recursion formulas:

dp[n][0] = dp[n-1][1]
dp[n][k] = dp[n-1][k-1] + dp[n-1][k+1] (1 <= k <= 8)
dp[n][9] = dp[n-1][8]

Step 4: Calculating the Final Result

The N-digit stair numbers can be found by summing up the cases for all digits from 0 to 9. The final result is calculated as sum(dp[N]).

Implementation

Now, let’s implement all of this logic in Python code:

def count_stair_numbers(n):
    # Constant for modular arithmetic
    MOD = 1000000000

    # Initialize DP table
    dp = [[0] * 10 for _ in range(n + 1)]

    # When the length is 1
    for i in range(1, 10):
        dp[1][i] = 1

    # Fill the DP table
    for i in range(2, n + 1):
        dp[i][0] = dp[i - 1][1] % MOD
        for j in range(1, 9):
            dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j + 1]) % MOD
        dp[i][9] = dp[i - 1][8] % MOD

    # Calculate the result
    result = sum(dp[n]) % MOD
    return result

# User Input
n = int(input("Enter the value of N: "))
print(count_stair_numbers(n))

Conclusion

Today, through the “Counting Stair Numbers” problem, we have understood the basic concepts of dynamic programming and explored the process of solving problems using Python code. I hope this will enhance your algorithm problem-solving skills. Stay tuned for the next tutorial where we will cover another interesting algorithm problem!

Problem Description

This is a problem of finding a path from the starting point to the endpoint in a given 2D array. This array consists of passages and obstacles, where passages are represented by 0 and obstacles by 1.

The goal is to determine whether there is a path from the given starting point to the endpoint and to output the possible path. Movement is allowed in four directions: up, down, left, and right.

Problem Definition

    def find_path(maze, start, end):
        """
        :param maze: 2D list representing the maze
        :param start: Tuple of (x, y) coordinates for the start point
        :param end: Tuple of (x, y) coordinates for the end point
        :return: List of tuples representing the path or None if no path exists
        """
    

Problem Example

    maze = [
        [0, 1, 0, 0, 0],
        [0, 1, 0, 1, 0],
        [0, 0, 0, 1, 0],
        [1, 1, 0, 0, 0],
        [0, 0, 0, 1, 1]
    ]
    start = (0, 0)  # Starting point
    end = (4, 4)    # Endpoint
    

Expected result:
[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (3, 2), (4, 2), (4, 3), (4, 4)]

Solution Approach

This problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS) algorithms. Here, I will explain how to find a path using DFS.

Depth-First Search (DFS) Algorithm

DFS explores all possible paths to determine if the endpoint is reachable. During the path search, it keeps track of visited nodes to prevent revisiting.

Implementation Steps

Step 1: Basic Setup

First, define the maze and the starting and ending points through basic setup.

    maze = [
        [0, 1, 0, 0, 0],
        [0, 1, 0, 1, 0],
        [0, 0, 0, 1, 0],
        [1, 1, 0, 0, 0],
        [0, 0, 0, 1, 1]
    ]
    start = (0, 0)
    end = (4, 4)
    

Step 2: Define the DFS Function

Implement a function to find the path using DFS.

    def dfs(maze, current, end, path, visited):
        if current == end:
            return path
        
        x, y = current
        visited.add(current)

        for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
            next_x, next_y = x + dx, y + dy
            
            if 0 <= next_x < len(maze) and 0 <= next_y < len(maze[0]):
                if maze[next_x][next_y] == 0 and (next_x, next_y) not in visited:
                    result = dfs(maze, (next_x, next_y), end, path + [(next_x, next_y)], visited)
                    if result is not None:
                        return result
        
        visited.remove(current)
        return None
    

Step 3: Implement the Path Finding Function

Finally, implement the path finding function to call DFS.

    def find_path(maze, start, end):
        visited = set()
        return dfs(maze, start, end, [start], visited)
    

Step 4: Output the Result

Call the function to output the result.

    path = find_path(maze, start, end)
    print("Path:", path)
    

Complete Code

    def dfs(maze, current, end, path, visited):
        if current == end:
            return path
        
        x, y = current
        visited.add(current)

        for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
            next_x, next_y = x + dx, y + dy
            
            if 0 <= next_x < len(maze) and 0 <= next_y < len(maze[0]):
                if maze[next_x][next_y] == 0 and (next_x, next_y) not in visited:
                    result = dfs(maze, (next_x, next_y), end, path + [(next_x, next_y)], visited)
                    if result is not None:
                        return result
        
        visited.remove(current)
        return None

    def find_path(maze, start, end):
        visited = set()
        return dfs(maze, start, end, [start], visited)

    maze = [
        [0, 1, 0, 0, 0],
        [0, 1, 0, 1, 0],
        [0, 0, 0, 1, 0],
        [1, 1, 0, 0, 0],
        [0, 0, 0, 1, 1]
    ]
    start = (0, 0)
    end = (4, 4)
    path = find_path(maze, start, end)
    print("Path:", path)
    

Conclusion

In this article, we examined a method using the DFS algorithm to solve the problem of finding a path in a 2D array. The code was structured to explore paths through DFS at each step and return results when conditions are met. Such basic pathfinding problems can be utilized as a foundation to tackle more complex problems.