Category: Swift Coding Test

Swift Coding Test Course, Interval Sum Calculation 1

Introduction

In today’s lecture, we will delve deep into the problem of calculating range sums. This problem frequently appears in various algorithmic challenges, and it’s essential to know efficient solutions.
In this article, we will define the problem, explain various approaches, and detail the process of finding the optimal solution.

Problem Definition

Given an array A and two integers L and R, calculate the value of
A[L] + A[L+1] + ... + A[R].
Assume that the index of A starts from 1.

Input

  • First line: Integer N (1 ≤ N ≤ 100,000) – Size of the array
  • Second line: N integers A[1], A[2], ..., A[N] (-1,000,000 ≤ A[i] ≤ 1,000,000)
  • Third line: Integer Q (1 ≤ Q ≤ 100,000) – Number of queries
  • Next Q lines: Each query contains two integers L and R

Output

For each query, output the range sum result line by line.

Problem Approach

There are several methods to solve this problem. A straightforward approach is to compute the sum directly for each query, but
this can be inefficient in terms of time complexity. Therefore, we will consider the following approach.

1. Approach through Simple Iteration

The most basic method is to use a loop to calculate the range sum for each query directly.
The time complexity of this method is O(N), and if there are Q queries, it becomes O(N*Q).
This would require up to a billion calculations in the worst-case scenario when both N and Q are at their maximum of 100,000, which is impractical.

2. Using a Cumulative Sum Array

Thus, using a cumulative sum array to solve the problem is much more efficient. With this approach,
the range sum can be resolved in O(1) time complexity. By creating a cumulative sum array and preprocessing the data linearly,
we can obtain results in O(1) time for each query.

Definition of Cumulative Sum Array

We will define the array p as follows:
p[i] = A[1] + A[2] + ... + A[i]
This way, the range sum A[L] + A[L+1] + ... + A[R] can be easily calculated as
p[R] - p[L-1].

Implementation

Now, let’s proceed with the actual implementation. I will write the algorithm using Swift. 


    import Foundation

    // Input
    let n = Int(readLine()!)!
    let a = readLine()!.split(separator: " ").map { Int($0)! }
    let q = Int(readLine()!)!

    // Initialize cumulative sum array
    var p = [0] + a
    
    // Create cumulative sum array
    for i in 1..

Results and Analysis

The above code constructs the cumulative sum array in O(N) time complexity and
prints the answer for each query in O(1) time.
In the worst-case scenario, considering the time spent on input, the overall time complexity is O(N + Q).

Advantages of the Optimized Approach

This method shows excellent performance, especially with large input data.
The use of cumulative sums allows for efficient handling of numerous queries.
Such problem-solving methods can be applied to other challenges and require a basic understanding of data structures.

Conclusion

Today, we explored an efficient problem-solving method using cumulative sum arrays through the problem of calculating range sums.
Many algorithmic problems practically utilize such techniques, so it is essential to grasp them.
In the next lecture, we will cover similar problems.
I hope this is helpful for your coding test preparation.

Swift Coding Test Course, Interval Sum

Problem Description

Given an integer array nums and two integers start and end, this is a problem to calculate the value of nums[start] + nums[start + 1] + ... + nums[end]. We will explore methods to efficiently calculate the range sum and how to solve this problem in Swift.

Input Example

[1, 2, 3, 4, 5], start = 1, end = 3

Output Example

9

Approach to the Problem

To solve this problem, we need to consider how to perform range addition efficiently. A simple way is to use a for loop to calculate the sum over the given index range. However, there is much room for improvement with this approach.

1. Sum Calculation Using Simple Loop

First, let’s take a look at the most basic method. This involves using a loop to calculate the sum for the given range. Below is the Swift code implementing this method.

func rangeSum(nums: [Int], start: Int, end: Int) -> Int {
        var sum = 0
        for i in start...end {
            sum += nums[i]
        }
        return sum
    }

Usage Example

let nums = [1, 2, 3, 4, 5]
let result = rangeSum(nums: nums, start: 1, end: 3)
print(result) // 9

2. Approach Using Cumulative Sum Array

A simple loop can lead to performance degradation in certain cases. Particularly, if we need to calculate the range sum multiple times for a large array, executing a loop each time is inefficient. In such cases, using a cumulative sum array is an effective approach.

By using a cumulative sum array, we can calculate the sum of a specific range of the array in constant time O(1). The approach is as follows:

  1. Create a cumulative sum array of the same size as the input array.
  2. Add the cumulative sum of the previous index to each index of the cumulative sum array.
  3. To calculate the range sum, we can easily compute it using prefixSum[end + 1] - prefixSum[start].

Cumulative Sum Array Implementation Code

func rangeSumUsingPrefix(nums: [Int], start: Int, end: Int) -> Int {
        var prefixSum = [0]    // Initialize cumulative sum array
        prefixSum.append(0)    // Initialize the first index to 0
        
        // Generate cumulative sum array
        for num in nums {
            prefixSum.append(prefixSum.last! + num)
        }
        
        // Calculate range sum
        return prefixSum[end + 1] - prefixSum[start]
    }

Usage Example

let nums = [1, 2, 3, 4, 5]
let result = rangeSumUsingPrefix(nums: nums, start: 1, end: 3)
print(result) // 9

Case Analysis

In this lecture, we have examined two approaches to solve the range sum problem. The method using a simple loop is intuitive and easy to understand, but can lead to performance degradation for large arrays. On the other hand, the method utilizing cumulative sum arrays is superior in terms of performance.

Conclusion

The range sum problem is a great example of utilizing algorithms and data structures. We learned that efficient approaches can lead to quicker solutions to problems. Try solving such problems using Swift and familiarize yourself with various algorithmic techniques.

References

Swift Coding Test Course, Finding the Number of Steps

Hello everyone! Today, we will tackle one of the frequently appearing problems in coding tests, the ‘Counting Stair Numbers‘ problem. In this article, we will explore the definition of the stair number problem, the solution methods, and step-by-step Swift code examples. Ultimately, we will include efficient and concise code implementations along with various test cases.

Problem Definition

A stair number (n) is a number of length n, meaning that the difference between two consecutive digits is 1. For example, 123, 234, and 321 are stair numbers. However, 122 and 3456 are not stair numbers. Your task is to calculate the count of n-digit stair numbers for a given n.

Examples of the Problem

Let’s take an example of finding stair numbers with a given number of digits (n) using the digits from 1 to 9. Here are some examples:

  • n = 1: Result = 9 (1, 2, 3, 4, 5, 6, 7, 8, 9)
  • n = 2: Result = 17 (10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 90)
  • n = 3: Result = 32 (101, 121, 123, 210, 212, …)

Approach to Solve the Problem

To solve this problem, we can use the Dynamic Programming technique. We solve the problem by utilizing the relationship between one-digit numbers and numbers above it, based on the properties of stair numbers.

Dynamic Programming Approach

First, we set the length of the stair number to n and establish a dp array to store the possible counts. dp[i][j] represents the count of stair numbers of length i with the last digit j. This allows us to set up a recurrence relation:

        dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1]

Here, j takes on values from 0 to 9. If the last digit is 0, the previous digit can only be 1, and if the last digit is 9, the previous digit can only be 8. Note that intermediate values can go both ways.

Basic Setup

Now, let’s initialize the DP table for solving the problem. Since the first digit of the stair numbers can only be from 1 to 9:

    for j in 0...9 {
        dp[1][j] = 1
    }

Next, we will set up a loop from the second digit up to n digits to fill the dp array.

Swift Code Implementation

Now, based on what we’ve explained above, let’s implement the code in Swift.

    func countStairNumbers(n: Int) -> Int {
        // Array to store the count of stair numbers for each digit
        var dp = Array(repeating: Array(repeating: 0, count: 10), count: n + 1)

        // For 1-digit numbers, there is 1 case for each from 1 to 9
        for j in 1...9 {
            dp[1][j] = 1
        }

        // Fill the DP array
        for i in 2...n {
            for j in 0...9 {
                if j > 0 {
                    dp[i][j] += dp[i - 1][j - 1]
                }
                if j < 9 {
                    dp[i][j] += dp[i - 1][j + 1]
                }
            }
        }

        var totalCount = 0
        // Sum up all cases for n-digit numbers
        for j in 0...9 {
            totalCount += dp[n][j]
        }

        return totalCount
    }

    // Usage example
    let result = countStairNumbers(n: 3)
    print("Count of 3-digit stair numbers: \(result)") // Output: 32

Code Explanation

The above code returns the count of n-digit stair numbers for a given n. The function sets up the dp array and calculates each possible case by adding the possibilities for each digit. Ultimately, it returns the total count of all n-digit numbers.

Test Cases

Let's create a few test cases to verify the correct operation of the function by checking the results for each digit count:

    print("1-digit count: \(countStairNumbers(n: 1))") // 9
    print("2-digit count: \(countStairNumbers(n: 2))") // 17
    print("3-digit count: \(countStairNumbers(n: 3))") // 32
    print("4-digit count: \(countStairNumbers(n: 4))") // X (to be computed and verified)
    print("5-digit count: \(countStairNumbers(n: 5))") // Y (to be computed and verified)

Conclusion

Today, we learned how to solve the counting stair numbers problem using Swift. We were able to devise an efficient solution to this problem through dynamic programming. I encourage you to understand and utilize this problem to tackle more algorithmic challenges!

Then, we'll see you next time with more interesting topics. Thank you!

Swift Coding Test Course, Finding the Product of an Interval

One of the common problems encountered in programming interviews or coding tests is ‘Calculating the Product of an Interval’. This problem requires a deep understanding of data structures and algorithms through the efficient calculation of the product over a specific interval of a given array. In this chapter, we will understand this problem and explore efficient solutions.

Problem Definition

Given an integer array nums and multiple queries queries, the task is to return the product of the interval defined by each query. The product of an interval refers to the result of multiplying all the numbers within that interval.

Example Problem

Example: 
Input: nums = [1, 2, 3, 4, 5]
Queries: queries = [[0, 1], [1, 3], [2, 4]]
Output: [2, 24, 60]
Explanation:
- The first query [0, 1] returns nums[0] * nums[1] = 1 * 2 = 2.
- The second query [1, 3] returns nums[1] * nums[2] * nums[3] = 2 * 3 * 4 = 24.
- The third query [2, 4] returns nums[2] * nums[3] * nums[4] = 3 * 4 * 5 = 60.

Solution Approach

The problem of calculating the product of an interval can be solved using a basic double loop. However, if the number of queries is large and the array is big, a double loop is inefficient, and we should consider other methods. Particularly, if the number of queries is N and each query takes O(M) time to calculate the interval product, the worst-case time complexity is O(N * M). Therefore, we need to find a method that can solve it in O(N + Q) time.

Calculating Interval Products in Linear Time Complexity

We can solve this problem using a technique called prefix product (pre-computation). This method helps to quickly calculate the product of an interval using the starting and ending indices of each query.

Calculating Prefix Product

First, we construct a prefix product array. The prefix product array is defined with the name prefixProduct, where prefixProduct[i] represents nums[0] * nums[1] * ... * nums[i]. This way, the product of the interval [i, j] can be simply calculated as prefixProduct[j] / prefixProduct[i-1].

Swift Code Implementation


import Foundation

func rangeProduct(_ nums: [Int], _ queries: [[Int]]) -> [Int] {
    var prefixProduct = [Int](repeating: 1, count: nums.count + 1)
    
    // Create the prefix product array
    for i in 0..

Code Analysis

The code above works through the following procedures:

  1. Creating the Prefix Product Array: It iterates through the nums array to calculate the cumulative product at each position. The time complexity of this process is O(N).
  2. Processing Queries: It checks each given query one by one and calculates the interval product. The time complexity for processing each query is O(1).
  3. Therefore, the overall time complexity is O(N + Q).

Conclusion

Through this tutorial, we have learned how to efficiently solve the problem of calculating the product of an interval. Moving away from the basic double loop approach to utilizing prefix products is a method that can be applied beneficially in many algorithmic problems. Such problems are also used in various applications for data processing in reality, so it is important to understand and utilize this problem well.

I wish you success in taking on more coding challenges and building your skills!

Swift Coding Test Course, Pathfinding

Hello, in this lecture, we will learn how to solve the pathfinding problem using the Swift language. Pathfinding is one of the common types of problems that often appear in coding tests, where the task is to find a path between two specific nodes in a given graph. In this article, we will define the problem and thoroughly discuss the algorithms and data structures needed to solve it.

Problem Definition

Let’s consider the following problem:

Problem: Write a function to determine whether there exists a path between two nodes A and B in a given directed graph. The graph is provided in the form of an adjacency list, with nodes and edges represented as integers.

Input:

  • Integer N: the number of nodes (1 ≤ N ≤ 1000)
  • Integer M: the number of edges (1 ≤ M ≤ 10000)
  • Length of the list M: each edge indicates a direction from node u to node v in the form [u, v].
  • Integer A, B: the starting node A and the ending node B to check for a path.

Output:

  • If there exists a path from node A to node B, print “YES”; if not, print “NO”.

Problem Analysis

This problem falls under the category of ‘path search’ in graph theory and can be solved using various methods. You can explore paths using DFS (Depth-First Search) or BFS (Breadth-First Search), both of which can effectively explore graphs in the form of adjacency lists.

Algorithm Selection

In this lecture, we will use BFS to explore the graph. BFS is an algorithm that uses a queue to explore each node level by level, making it advantageous for finding the shortest path. We will check if we can reach the destination node through graph traversal.

Swift Code Implementation

Now, let’s write the actual Swift code. Below is an example of a pathfinding function that uses BFS.

        
        import Foundation

        func canReach(start: Int, end: Int, edges: [[Int]], nodeCount: Int) -> String {
            var adjacencyList = [[Int]](repeating: [Int](), count: nodeCount + 1)
            for edge in edges {
                adjacencyList[edge[0]].append(edge[1])
            }

            var queue = [start]
            var visited = [Bool](repeating: false, count: nodeCount + 1)
            visited[start] = true

            while !queue.isEmpty {
                let current = queue.removeFirst()

                if current == end {
                    return "YES"
                }

                for neighbor in adjacencyList[current] {
                    if !visited[neighbor] {
                        visited[neighbor] = true
                        queue.append(neighbor)
                    }
                }
            }
            return "NO"
        }

        // Example input
        let nodeCount = 6
        let edges = [[1, 2], [1, 3], [2, 4], [4, 5], [3, 6]]
        let start = 1
        let end = 5

        print(canReach(start: start, end: end, edges: edges, nodeCount: nodeCount))

Code Explanation

Analyzing the above code, it consists of the following steps:

  1. Create a graph in the form of an adjacency list. Store connected nodes for each node.
  2. Initialize a queue for BFS traversal and mark the starting node as visited.
  3. Repeat the following process until the queue is empty:
    • Take a node from the front of the queue.
    • If the taken node is the destination node, return “YES”.
    • For all adjacent nodes of the current node, add unvisited nodes to the queue and mark them as visited.
  4. If the queue is empty but the target node has not been reached, return “NO”.

Complexity Analysis

The time complexity of the BFS algorithm is O(V + E), where V is the number of nodes and E is the number of edges. Considering the maximum conditions given in this problem, it is very efficient. The memory complexity also requires O(V + E) to store the adjacency list.

Testing and Applications

The given function can be applied to various graph structures and tested accordingly. Let’s look at a few additional test cases.

        
        // Additional test cases
        let additionalEdges1 = [[1, 2], [2, 3], [3, 4], [4, 5]]
        let additionalEdges2 = [[1, 2], [2, 3], [3, 4], [2, 5]]
        
        print(canReach(start: 1, end: 5, edges: additionalEdges1, nodeCount: nodeCount)) // NO
        print(canReach(start: 1, end: 5, edges: additionalEdges2, nodeCount: nodeCount)) // YES

Conclusion

In this lecture, we learned how to solve the pathfinding problem in graphs using Swift. We effectively solved the problem using the BFS algorithm, which is very important in coding tests. When encountering similar problems in practice, you will be able to solve them based on the knowledge gained from this lecture.

Now it’s your turn to tackle various graph problems and enhance your algorithm skills. Keep practicing!