To help you prepare for coding tests, we introduce an algorithm problem that can be solved using the Swift language. This problem is ‘Finding the Desired Integer’, where you will implement an algorithm to find a specific integer within a given integer array. This article will explain the process from problem description to algorithm solution in detail.<\/p>\n<\/header>\n You are given an integer array If To solve this problem, one must think of how to find the index of Linear search is a method where you check each element one by one from the beginning to the end of the array to find 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 First, we will implement linear search and then also implement binary search.<\/p>\n We will run a few cases to test the function we have written.<\/p>\nProblem Description<\/h2>\n
numbers<\/code>. Write a function to check if a specific integer target<\/code> exists in this array. If it exists, return the index of that integer; if not, return -1<\/code>.<\/p>\nfunc findTarget(numbers: [Int], target: Int) -> Int<\/code><\/pre>\n<\/section>\nInput<\/h2>\n
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numbers<\/code>: An array of integers (1 \u2264 length of numbers \u2264 106<\/sup>)<\/li>\ntarget<\/code>: The integer you want to find (-109<\/sup> \u2264 target \u2264 109<\/sup>)<\/li>\n<\/ul>\n<\/section>\nOutput<\/h2>\n
target<\/code> exists, return the corresponding index; otherwise, return -1<\/code>.<\/p>\n<\/section>\nExamples<\/h2>\n
Example 1<\/h3>\n
Input: numbers = [1, 2, 3, 4, 5], target = 3\nOutput: 2<\/code><\/pre>\nExample 2<\/h3>\n
Input: numbers = [5, 6, 7, 8, 9], target = 1\nOutput: -1<\/code><\/pre>\n<\/section>\nSolution Approach<\/h2>\n
target<\/code> within the array. A basic method is linear search using a loop, and if the array is sorted, binary search can be used.<\/p>\n1. Linear Search<\/h3>\n
target<\/code>. The time complexity is O(n) depending on the length of the array.<\/p>\n2. Binary Search<\/h3>\n
target<\/code>. The time complexity in this case is O(log n).<\/p>\n<\/section>\nImplementation<\/h2>\n
Linear Search Implementation<\/h3>\n
func findTarget(numbers: [Int], target: Int) -> Int {\n for (index, number) in numbers.enumerated() {\n if number == target {\n return index\n }\n }\n return -1\n}<\/code><\/pre>\nBinary Search Implementation<\/h3>\n
func binarySearch(numbers: [Int], target: Int) -> Int {\n var left = 0\n var right = numbers.count - 1\n\n while left <= right {\n let mid = (left + right) \/ 2\n if numbers[mid] == target {\n return mid\n } else if numbers[mid] < target {\n left = mid + 1\n } else {\n right = mid - 1\n }\n }\n return -1\n}<\/code><\/pre>\n<\/section>\nTest Cases<\/h2>\n
let numbers = [1, 2, 3, 4, 5]\nlet target1 = 3\nlet target2 = 6\n\nprint(findTarget(numbers: numbers, target: target1)) \/\/ 2\nprint(findTarget(numbers: numbers, target: target2)) \/\/ -1<\/code><\/pre>\n<\/section>\nPerformance Evaluation<\/h2>\n