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class Solution:
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def countAndSay(self, n: int) -> str:
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"""
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Complexities:
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Time: O(n2^n)
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Space: O()
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"""
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# O(len(s)) where len(s) ~ O(2^n)
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def RLE(s: str) -> str:
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rle = ""
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checkChar = s[0]
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charCount = 0
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for i, l in enumerate(s):
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if checkChar == l:
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charCount += 1
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else:
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rle += f"{charCount}{checkChar}"
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checkChar = l
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charCount = 1
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if checkChar:
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rle += f"{charCount}{checkChar}"
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return rle
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lastSol = "1"
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# Iteratively call RLE on lastSol (equiv. to countAndSay(n-1))
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# - O(n2^n) = n * O(2^n)
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for i in range(2, n + 1):
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lastSol = RLE(lastSol)
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return lastSol
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