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뉴스 클러스터링 #36

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37 changes: 37 additions & 0 deletions WEEK03/프로그래머스_[1차] 뉴스 클러스터링/JSON.swift
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Original file line number Diff line number Diff line change
@@ -0,0 +1,37 @@
func solution(_ str1:String, _ str2:String) -> Int {
var result = 65536
let (lhs, rhs) = (parse(str1), parse(str2))

let intersection = getIntersection(lhs, rhs)
let union = lhs.values.reduce(0, +) + rhs.values.reduce(0, +) - intersection

guard union != 0 else { return 65536 }

result = Int(Double(intersection) / Double(union) * 65536)

return result
Comment on lines +10 to +12
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@gemini-code-assist gemini-code-assist bot Feb 8, 2026

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medium

result 변수에 값을 할당한 후 바로 반환하는 대신, 계산된 값을 직접 반환하여 코드를 간결하게 만들 수 있습니다. 이 변경으로 인해 2행의 var result = 65536 선언이 불필요해지므로 함께 제거하는 것이 좋습니다. Swift에서는 불필요한 가변 상태를 줄이는 것이 권장됩니다.

Suggested change
result = Int(Double(intersection) / Double(union) * 65536)
return result
return Int(Double(intersection) / Double(union) * 65536)

}

func getIntersection(_ lhs: [String: Int], _ rhs: [String: Int]) -> Int {
var result = 0

for word in lhs.keys {
result += min(lhs[word, default: 0], rhs[word, default: 0])
}

return result
Comment on lines +16 to +22
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@gemini-code-assist gemini-code-assist bot Feb 8, 2026

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medium

for 루프를 사용하여 합계를 계산하는 대신 reduce 고차 함수를 사용하면 코드를 더 간결하고 함수형 프로그래밍 스타일에 가깝게 작성할 수 있습니다.

    return lhs.keys.reduce(0) { result, word in
        result + min(lhs[word, default: 0], rhs[word, default: 0])
    }

}

func parse(_ str: String) -> [String: Int] {
var result = [String: Int]()
let arr = Array(str)

for i in 0..<str.count - 1 {
guard arr[i].isLetter && arr[i + 1].isLetter else { continue }

let word = String(arr[i]) + String(arr[i + 1])
result[word.lowercased(), default: 0] += 1
}

return result
Comment on lines +26 to +36
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medium

parse 함수 내에서 루프가 반복될 때마다 lowercased()가 호출되고 있습니다. 입력 문자열을 루프 시작 전에 한 번만 소문자로 변환하면 성능을 개선할 수 있습니다. 또한, 두 문자를 합쳐 문자열을 만들 때 String의 이니셜라이저에 배열 슬라이스를 전달하는 것이 문자열 덧셈보다 약간 더 효율적입니다.

    var result = [String: Int]()
    let arr = Array(str.lowercased())

    for i in 0..<(arr.count - 1) {
        guard arr[i].isLetter && arr[i + 1].isLetter else { continue }
        
        let word = String(arr[i...i+1])
        result[word, default: 0] += 1
    }

    return result

}