LeetCode 周赛 440

392
Algorithm
LeetCodeContest
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Q1. 将水果放入篮子 Ⅱ ​

解题思路 ​

暴力

示例代码 ​

python
class Solution:
    def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
        ans = len(fruits)
        for i, x in enumerate(fruits):
            for j, y in enumerate(baskets):
                if y >= x:
                    baskets[j] = 0
                    ans -= 1
                    break
        return ans

Q2. 选出和最大的 K 个元素 ​

解题思路 ​

用堆保留 K 个最大的元素

示例代码 ​

python
class Solution:
    def findMaxSum(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:
        n = len(nums1)
        s = i = 0
        h = []
        ans = [0] * n
        li = sorted([(x, i) for i, x in enumerate(nums1)])
        while i < n:
            while len(h) > k:
                s -= heappop(h)
            st = i
            tmp = s
            while i < n and li[i][0] == li[st][0]:
                heappush(h, nums2[li[i][1]])
                s += nums2[li[i][1]]
                ans[li[i][1]] = tmp
                i += 1
        return ans

Q3. 将水果放入篮子 Ⅲ ​

解题思路 ​

参考：线段树二分（Python/Java/C++/Go）

示例代码 ​

python
class Solution:
  def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
    n = len(fruits)
    ans = 0
    segment_tree = SegTree(lambda x, y: x if x > y else y, -1, baskets)

    for i, x in enumerate(fruits):
      i = segment_tree.max_right(0, lambda y: y < x)
      if i == n:
        ans += 1
      else:
        segment_tree.set(i, -1)
    return ans
py
import typing


class SegTree:
    __slots__ = ["_op", "_e", "_n", "_log", "_size", "_d"]

    def __init__(self, op: typing.Callable[[typing.Any, typing.Any], typing.Any],
                       e: typing.Any,
                       v: typing.Union[int, typing.List[typing.Any]]) -> None:
        """
        Args:
                op: maintain operation function, such as add, max, min, etc.
                e: initial value of the segment tree
                v: initial value of the array
        """
        self._op = op
        self._e = e

        if isinstance(v, int):
            v = [e] * v

        self._n = len(v)
        self._log = (self._n - 1).bit_length()
        self._size = 1 << self._log
        self._d = [e] * (2 * self._size)

        for i in range(self._n):
            self._d[self._size + i] = v[i]
        for i in range(self._size - 1, 0, -1):
            self._update(i)

    def set(self, p: int, x: typing.Any) -> None:
        assert 0 <= p < self._n

        p += self._size
        self._d[p] = x
        for i in range(1, self._log + 1):
            self._update(p >> i)

    def get(self, p: int) -> typing.Any:
        assert 0 <= p < self._n

        return self._d[p + self._size]

    def prod(self, left: int, right: int) -> typing.Any:
        assert 0 <= left <= right <= self._n

        sml = self._e
        smr = self._e
        left += self._size
        right += self._size

        while left < right:
            if left & 1:
                sml = self._op(sml, self._d[left])
                left += 1
            if right & 1:
                right -= 1
                smr = self._op(self._d[right], smr)
            left >>= 1
            right >>= 1

        return self._op(sml, smr)

    def all_prod(self) -> typing.Any:
        return self._d[1]

    def max_right(self, left: int,
                                f: typing.Callable[[typing.Any], bool]) -> int:
        assert 0 <= left <= self._n
        assert f(self._e)

        if left == self._n:
            return self._n

        left += self._size
        sm = self._e

        first = True
        while first or (left & -left) != left:
            first = False
            while left % 2 == 0:
                left >>= 1
            if not f(self._op(sm, self._d[left])):
                while left < self._size:
                    left *= 2
                    if f(self._op(sm, self._d[left])):
                        sm = self._op(sm, self._d[left])
                        left += 1
                return left - self._size
            sm = self._op(sm, self._d[left])
            left += 1

        return self._n

    def min_left(self, right: int,
                             f: typing.Callable[[typing.Any], bool]) -> int:
        assert 0 <= right <= self._n
        assert f(self._e)

        if right == 0:
            return 0

        right += self._size
        sm = self._e

        first = True
        while first or (right & -right) != right:
            first = False
            right -= 1
            while right > 1 and right % 2:
                right >>= 1
            if not f(self._op(self._d[right], sm)):
                while right < self._size:
                    right = 2 * right + 1
                    if f(self._op(self._d[right], sm)):
                        sm = self._op(self._d[right], sm)
                        right -= 1
                return right + 1 - self._size
            sm = self._op(self._d[right], sm)

        return 0

    def _update(self, k: int) -> None:
        self._d[k] = self._op(self._d[2 * k], self._d[2 * k + 1])

Q4. 删除一个冲突对后最大字数组数目 ​

这道题很神秘，以后再来解决吧！

Inverse
Collections