😎 κ³΅λΆ€ν•˜λŠ” μ§•μ§•μ•ŒνŒŒμΉ΄λŠ” μ²˜μŒμ΄μ§€?

[Programmers] μ˜μƒ λ³Έλ¬Έ

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[Programmers] μ˜μƒ

μ§•μ§•μ•ŒνŒŒμΉ΄ 2023. 5. 5. 23:23
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λ°˜μ‘ν˜•
μ½”λ‹ˆλŠ” λ§€μΌ λ‹€λ₯Έ μ˜·μ„ μ‘°ν•©ν•˜μ—¬ μž…λŠ”κ²ƒμ„ μ’‹μ•„함
각 μ’…λ₯˜λ³„λ‘œ μ΅œλŒ€ 1가지 μ˜μƒλ§Œ μ°©μš©
μ°©μš©ν•œ μ˜μƒμ˜ μΌλΆ€κ°€ κ²ΉμΉ˜λ”라도, λ‹€λ₯Έ μ˜μƒμ΄ κ²ΉμΉ˜μ§€ μ•Šκ±°λ‚˜, ν˜Ήμ€ μ˜μƒμ„ μΆ”κ°€λ‘œ λ” μ°©μš©ν•œ κ²½μš°μ—λŠ” μ„œλ‘œ λ‹€λ₯Έ λ°©λ²•μœΌλ‘œ μ˜·μ„ μ°©μš©ν•œ κ²ƒμœΌλ‘œ κ³„μ‚°
μ½”λ‹ˆλŠ” ν•˜λ£¨μ— μ΅œμ†Œ ν•œ κ°œμ˜ μ˜μƒμ€ μž…μŒ

μ½”λ‹ˆκ°€ κ°€μ§„ μ˜μƒλ“€μ΄ λ‹΄κΈ΄ 2차원 λ°°μ—΄ clothesκ°€ μ£Όμ–΄μ§ˆ λ•Œ μ„œλ‘œ λ‹€λ₯Έ μ˜·μ˜ μ‘°ν•©μ˜ μˆ˜λ₯Ό return
  clothes의 κ° ν–‰μ€ [μ˜μƒμ˜ μ΄λ¦„, μ˜μƒμ˜ μ’…λ₯˜]
  μ½”λ‹ˆκ°€ κ°€μ§„ μ˜μƒμ˜ μˆ˜λŠ” 1개 μ΄μƒ 30개 μ΄ν•˜
  κ°™μ€ μ΄λ¦„을 κ°€μ§„ μ˜μƒμ€ μ‘΄μž¬ν•˜μ§€ μ•ŠμŒ
  clothes의 λͺ¨λ“  μ›μ†ŒλŠ” λ¬Έμžμ—΄
  λͺ¨λ“  λ¬Έμžμ—΄μ˜ κΈΈμ΄λŠ” 1 μ΄μƒ 20 μ΄ν•˜μΈ μžμ—°μˆ˜, μ•ŒνŒŒλ²³ μ†Œλ¬Έμž λ˜λŠ” '_'
def solution(clothes):
    closet = {}
    answer = 1
    for i in clothes:
        # μ’…λ₯˜
        key = i[1]
        value = i[0]
        if key in closet:
            closet[key].append(value)
        else:
            closet[key] = [value]

    for key in closet.keys():
        # λͺ¨λ“  경우의 수 - μ•„μ˜ˆ μ•ˆμž…μ€ κ±°
        answer = answer * (len(closet[key]) + 1)
    return answer - 1
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