{"id":34872,"date":"2024-11-01T09:32:59","date_gmt":"2024-11-01T09:32:59","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34872"},"modified":"2024-11-01T11:26:10","modified_gmt":"2024-11-01T11:26:10","slug":"swift-coding-test-course-finding-parenthesis-arrangement-to-make-minimum-value","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34872\/","title":{"rendered":"Swift Coding Test Course, Finding Parenthesis Arrangement to Make Minimum Value"},"content":{"rendered":"<p><body><\/p>\n<h2>Problem: Finding Parentheses Arrangement that Minimizes Value<\/h2>\n<p>Algorithm problems play an important role in coding tests, and especially problems related to parentheses are often presented. In this article, we will address the challenge of finding the minimum value by considering all possible parentheses arrangements for a given expression. We will explain the theories and algorithms necessary to solve this problem in detail.<\/p>\n<h3>Problem Description<\/h3>\n<p>There is a string composed of given numbers and operators. This string can be calculated by placing parentheses in various ways. Our goal is to find the minimum value that can be obtained by placing parentheses.<\/p>\n<p><strong>Input<\/strong>: &#8220;1+2*3-4*5&#8221; <br \/>\n<strong>Output<\/strong>: -1<\/p>\n<h3>Approach to the Problem<\/h3>\n<p>This problem can be solved using Dynamic Programming. We need to consider all possible placements of parentheses for the given expression, so the Divide and Conquer technique will also be utilized.<\/p>\n<h4>1. Parsing the Expression<\/h4>\n<p>First, we need to separate the numbers and operators from the expression and store them in arrays. For example, we will convert the string &#8220;1+2*3-4*5&#8221; into the following format.<\/p>\n<pre>\n    Number array: [1, 2, 3, 4, 5]\n    Operator array: ['+', '*', '-', '*']\n    <\/pre>\n<h4>2. Defining Dynamic Programming<\/h4>\n<p>Dynamic programming is a method that stores the results of each subproblem through several auxiliary arrays for future use. By using memoization, we can reduce time complexity by reusing results that have already been calculated.<\/p>\n<h4>3. Implementing the Recursive Function<\/h4>\n<p>We will calculate all possible combinations through a recursive function. This function divides based on the given index and calculates by placing parentheses differently for each side.<\/p>\n<h3>Swift Implementation<\/h3>\n<p>Below is the complete code implemented in the Swift programming language.<\/p>\n<pre><code>\n    func minValue(expression: String) -> Int {\n        var numbers: [Int] = []\n        var operators: [Character] = []\n        \n        \/\/ Step 1: Parse the expression\n        var currentNumber = \"\"\n        for char in expression {\n            if char.isNumber {\n                currentNumber.append(char)\n            } else {\n                numbers.append(Int(currentNumber)!)\n                operators.append(char)\n                currentNumber = \"\"\n            }\n        }\n        numbers.append(Int(currentNumber)!)\n        \n        \/\/ Step 2: Create a memoization table\n        var memo: [[Int?]] = Array(repeating: Array(repeating: nil, count: numbers.count), count: numbers.count)\n        \n        \/\/ Step 3: Define a recursive function\n        func compute(_ left: Int, _ right: Int, _ op: Character) -> Int {\n            switch op {\n            case '+': return left + right\n            case '-': return left - right\n            case '*': return left * right\n            default: fatalError(\"Unknown operator\")\n            }\n        }\n\n        \/\/ Recursive function for minimum value\n        func findMinValue(left: Int, right: Int) -> Int {\n            if left == right {\n                return numbers[left]\n            }\n            if let result = memo[left][right] {\n                return result\n            }\n\n            var result = Int.max\n            for i in left..<right {\n                let leftValue = findMinValue(left: left, right: i)\n                let rightValue = findMinValue(left: i + 1, right: right)\n                let computedValue = compute(leftValue, rightValue, operators[i])\n                result = min(result, computedValue)\n            }\n            memo[left][right] = result\n            return result\n        }\n\n        \/\/ Calculate the minimum value for the whole expression\n        return findMinValue(left: 0, right: numbers.count - 1)\n    }\n<\/code><\/pre>\n<h3>Time Complexity Analysis<\/h3>\n<p>The time complexity of this algorithm is O(N^3). N is the number of digits, as we need to evaluate the operators for each combination of two digits. However, due to the use of memoization, we do not recalculate the same subproblems repeatedly.<\/p>\n<h3>Conclusion<\/h3>\n<p>In this lecture, we covered the problem of \"Finding Parentheses Arrangement that Minimizes Value\" and explored the process of solving the problem and its implementation in Swift. I hope you achieve good results in coding tests through various approaches and optimization techniques to algorithm problems. Thank you!<\/p>\n<p><\/body><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problem: Finding Parentheses Arrangement that Minimizes Value Algorithm problems play an important role in coding tests, and especially problems related to parentheses are often presented. In this article, we will address the challenge of finding the minimum value by considering all possible parentheses arrangements for a given expression. We will explain the theories and algorithms &hellip; <a href=\"https:\/\/atmokpo.com\/w\/34872\/\" class=\"more-link\">\ub354 \ubcf4\uae30<span class=\"screen-reader-text\"> &#8220;Swift Coding Test Course, Finding Parenthesis Arrangement to Make Minimum Value&#8221;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[129],"tags":[],"class_list":["post-34872","post","type-post","status-publish","format-standard","hentry","category-swift-coding-test"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Swift Coding Test Course, Finding Parenthesis Arrangement to Make Minimum Value - \ub77c\uc774\ube0c\uc2a4\ub9c8\ud2b8<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/atmokpo.com\/w\/34872\/\" \/>\n<meta property=\"og:locale\" content=\"ko_KR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Swift Coding Test Course, Finding Parenthesis Arrangement to Make Minimum Value - \ub77c\uc774\ube0c\uc2a4\ub9c8\ud2b8\" \/>\n<meta property=\"og:description\" content=\"Problem: Finding Parentheses Arrangement that Minimizes Value Algorithm problems play an important role in coding tests, and especially problems related to parentheses are often presented. 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