Author: root

Problem Description

You are given an integer array numbers. Write a function to check if a specific integer target exists in this array. If it exists, return the index of that integer; if not, return -1.

func findTarget(numbers: [Int], target: Int) -> Int

Input

• numbers: An array of integers (1 ≤ length of numbers ≤ 10⁶)
• target: The integer you want to find (-10⁹ ≤ target ≤ 10⁹)

Output

If target exists, return the corresponding index; otherwise, return -1.

Examples

Example 1

Input: numbers = [1, 2, 3, 4, 5], target = 3
Output: 2

Example 2

Input: numbers = [5, 6, 7, 8, 9], target = 1
Output: -1

Solution Approach

To solve this problem, one must think of how to find the index of target within the array. A basic method is linear search using a loop, and if the array is sorted, binary search can be used.

1. Linear Search

Linear search is a method where you check each element one by one from the beginning to the end of the array to find target. The time complexity is O(n) depending on the length of the array.

2. Binary Search

Binary search is much more efficient when the array is sorted. This method reduces the search range by half based on the middle index of the array to find target. The time complexity in this case is O(log n).

Implementation

First, we will implement linear search and then also implement binary search.

Linear Search Implementation

func findTarget(numbers: [Int], target: Int) -> Int {
    for (index, number) in numbers.enumerated() {
        if number == target {
            return index
        }
    }
    return -1
}

Binary Search Implementation

func binarySearch(numbers: [Int], target: Int) -> Int {
    var left = 0
    var right = numbers.count - 1

    while left <= right {
        let mid = (left + right) / 2
        if numbers[mid] == target {
            return mid
        } else if numbers[mid] < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return -1
}

Test Cases

We will run a few cases to test the function we have written.

let numbers = [1, 2, 3, 4, 5]
let target1 = 3
let target2 = 6

print(findTarget(numbers: numbers, target: target1)) // 2
print(findTarget(numbers: numbers, target: target2)) // -1

Performance Evaluation

Linear search is simple but, in the worst case, requires searching through all elements, taking an average of O(n) time. In contrast, binary search performs better on sorted arrays and guarantees O(log n). Therefore, binary search is recommended for large datasets.

Conclusion

Through this problem, we learned how to find integers in an array using Swift. We understood how to solve the problem using basic linear search and also learned about improving performance with more efficient binary search. We encourage you to develop your skills in solving algorithm problems through continuous practice.

Problem Definition

The Travelling Salesman Problem (TSP) involves finding the shortest route for a salesman to visit n cities exactly once and return to the starting city, given the distances between each pair of cities. The inputs consist of the number of cities n and the distance matrix between those cities.

Input Format

• First line: Number of cities n (1 ≤ n ≤ 10)
• Next n lines: Distance matrix, where the i-th line contains the distances from the i-th city to each city (A[i][j] is the distance from city i to city j)

Output Format

Output the length of the shortest route that visits all cities and returns to the starting city.

Problem-Solving Approach

This problem belongs to the NP-Hard category, so it can be solved using dynamic programming techniques combined with pruning. This method allows us to only choose the least costly paths while constructing routes, rather than exploring all possible paths.

Step 1: Data Structure Design

First, declare a 2D array to store distance information between cities. Additionally, declare an array to check which cities have been visited and a variable to store the total distance so far.

Step 2: Using Bitmask and Recursive Functions

Use bitmasking to represent whether each city has been visited. Every time we move to the next city via a recursive function, check off the visited cities and calculate the distance of the path taken once all cities have been visited and the route returns to the starting city. Since visited cities can be easily represented with a bitmask, they can be processed efficiently.

Step 3: Finding the Optimal Route

Instead of exploring all possible paths, proceed by calculating the minimum distance among the currently selected paths. If the given path has visited all cities, compare it with the previous distance to update the minimum value.

Swift Code Implementation

Below is the code for finding the travelling salesman route using Swift.

let INF = Int.max
var dist: [[Int]] = []
var n = 0
var dp: [[Int]] = []
var visited: Int = 0
var ans: Int = INF

func tsp(pos: Int, visited: Int) -> Int {
    if visited == (1 << n) - 1 {
        return dist[pos][0] != 0 ? dist[pos][0] : INF
    }
    
    if dp[pos][visited] != -1 {
        return dp[pos][visited]
    }
    
    var minimum = INF
    for city in 0.. Int {
    visited = 1 // start from city 0
    dp = Array(repeating: Array(repeating: -1, count: 1 << n), count: n)
    let minimumDistance = tsp(pos: 0, visited: visited)
    return minimumDistance
}

// Example distance matrix
dist = [
    [0, 10, 15, 20],
    [10, 0, 35, 25],
    [15, 35, 0, 30],
    [20, 25, 30, 0]
]
n = dist.count

let result = findMinimumRoute()
print("The length of the minimum route is \(result).")

Code Analysis and Functionality

This code uses bitmasking to check visit completion and recursively explores possible routes to calculate the minimum route. Every time a city is visited, the total distance is updated, and it finally returns the distance of the route that goes back to the starting city after visiting all cities.

1. Initializing the dp Array

The dp array stores the minimum route distances based on each city and visit status. Initially, all elements are set to -1 to indicate an unvisited state.

2. Recursive Function tsp()

The recursive function tsp() takes the current position (pos) and the visited city bitmask (visited) as parameters to compute the optimal route. If all cities have been visited, it compares the cost of returning to the starting city and returns the minimum value.

3. Route Calculation

Bitmasking checks for cities that have not yet been visited and calculates the move to the next city, exploring all possibilities. The distances between cities are accumulated by calling `tsp()` for the entire route distance.

Conclusion

In this tutorial, we explored the Travelling Salesman Problem in detail. We discussed how to efficiently solve it using bitmasking and DP while writing a Swift code. I hope this problem, which frequently appears in coding tests, enhances your algorithm problem-solving skills.

I wish you the best in your coding journey! Thank you.

1. Problem Definition

The next greater element is the first number that is greater than a specific element, located to its right in a given sequence. If such a number does not exist, it returns -1. In other words, for a sequence a[0], a[1], ..., a[n-1], the goal is to find the smallest j satisfying a[j] (j > i) for each a[i].

For example, if the given array is [2, 1, 4, 3], the next greater elements are as follows:

• The next greater element for 2 is 4.
• The next greater element for 1 is 4.
• The next greater element for 4 is -1.
• The next greater element for 3 is -1.

As a result, the returned next greater element array will be [4, 4, -1, -1].

2. Problem Approach

To solve this problem, we will consider a stack-based approach. A stack is a LIFO (Last In First Out) structure, making it a very efficient data structure for inserting or deleting elements. The following procedure will be followed to find the next greater elements:

1. Initialize an empty stack.
2. Traverse the array from left to right.
3. Check the current element a[i] and:
1. If the stack is not empty and the number pointed to by the index at the top of the stack is less than a[i]:
• Set the next greater element for the index at the top of the stack to a[i].
• Remove that index from the stack.
2. Add the current index to the stack.
4. After traversing all elements of the array, set the next greater element to -1 for the indices remaining in the stack.

This method has a time complexity of O(n) because each element is added and removed from the stack only once. This is efficient, and as a result, it can deliver good performance even with large input values.

3. Swift Code Implementation

Now let’s implement the algorithm described above in Swift. Below is a Swift function to calculate the next greater elements:

import Foundation

func nextGreaterElement(_ nums: [Int]) -> [Int] {
    let n = nums.count
    var result = Array(repeating: -1, count: n) // Initialize result array
    var stack = [Int]() // Stack to store indices
    
    for i in 0..