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61 lines
2.2 KiB
Python
61 lines
2.2 KiB
Python
def count_beautiful_integers(low: int, high: int, k: int) -> int:
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""" Count the number of beautiful integers in the range [low, high].
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A number is beautiful if it meets both conditions:
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1. The count of even digits equals the count of odd digits
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2. The number is divisible by k
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Args:
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low: Lower bound of the range (inclusive), 0 < low <= high <= 10^9
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high: Upper bound of the range (inclusive)
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k: Divisor to check, 0 < k <= 20
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Returns:
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The count of beautiful integers in the given range
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>>> count_beautiful_integers(10, 20, 3)
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2
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>>> count_beautiful_integers(1, 10, 1)
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1
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>>> count_beautiful_integers(5, 5, 2)
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0
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"""
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from functools import lru_cache
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def count_upto(x: int) -> int:
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if x <= 0:
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return 0
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s = str(x)
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digits = list(map(int, s))
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n = len(digits)
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@lru_cache(maxsize=None)
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def dfs(pos: int, tight: bool, started: bool, diff: int, mod: int) -> int:
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# diff = (#even digits so far) - (#odd digits so far), valid only if started
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# mod = value so far % k, valid only if started
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if pos == n:
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if started and diff == 0 and mod == 0:
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return 1
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return 0
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limit = digits[pos] if tight else 9
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total = 0
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for d in range(limit + 1):
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ntight = tight and (d == limit and d == digits[pos])
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# Actually ntight should be tight and (d == digits[pos])
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# The above condition with limit is equivalent when tight,
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# but be precise:
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# (recompute for clarity)
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# ntight = tight and (d == digits[pos])
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nstarted = started or (d != 0)
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if not nstarted:
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total += dfs(pos + 1, ntight, False, 0, 0)
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else:
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ndiff = diff + (1 if (d % 2 == 0) else -1)
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nmod = (mod * 10 + d) % k
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total += dfs(pos + 1, ntight, True, ndiff, nmod)
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return total
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return dfs(0, True, False, 0, 0)
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return count_upto(high) - count_upto(low - 1)
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