力扣第 439 场周赛

509
Contest
LeetCode
本文最后更新于 2 分钟前，文中所描述的信息可能已发生改变。

Q1. 执行指令后的得分 ​

解题思路 ​

模拟题，根据题意进行遍历，累加答案

示例代码 ​

python
class Solution:
    def calculateScore(self, instructions: List[str], values: List[int]) -> int:
        n = len(instructions)
        vis = set()
        ans = 0
        i = 0
        while i not in vis and 0 <= i < n:
            vis.add(i)
            if instructions[i] == 'add':
                ans += values[i]
                i += 1
            else:
                i += values[i]
        return ans

Q2. 非递减数组的最大长度 ​

解题思路 ​

单调栈，维护区间的最大值

示例代码 ​

python
class Solution:
    def maximumPossibleSize(self, nums: List[int]) -> int:
        st = []
        for x in nums:
            cur = x
            while st and st[-1] > cur:
                cur = max(cur, st.pop())
            st.append(cur)
        return len(st)

Q3. 求出数组的 X 值 I ​

解题思路 ​

线性 DP，维护前一个数的余数状态，也就 k 个，总体时间复杂度为 O(nk)

示例代码 ​

python
class Solution:
    def resultArray(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        ans = [0] * k
        f = [0] * k
        for x in nums:
            x %= k
            nf = [0] * k
            nf[x] += 1
            for i in range(k):
                nf[(i * x) % k] += f[i]
            for i in range(k):
                ans[i] += nf[i]
            f = nf[:]
        return ans

Q4. 求出数组的 X 值 II ​

解题思路 ​

参考：从分治到线段树（Python/Java/C++/Go）

示例代码 ​

python
class Solution:
    def resultArray(self, nums: List[int], k: int, queries: List[List[int]]) -> List[int]:
        n = len(nums)
        nums = [x % k for x in nums]

        def op(a: Tuple[int, List[int]], b: Tuple[int, List[int]]) -> Tuple[int, List[int]]:
            cnt = a[1].copy()
            left_m = a[0]
            for m, c in enumerate(b[1]):
                cnt[left_m * m % k] += c
            return left_m * b[0] % k, cnt

        e = (1, [0] * k)
        v = []
        for x in nums:
            res = [0] * k
            res[x] += 1
            v.append((x, res))

        sg = SegTree(op, e, v)
        ans = []
        for i, v, s, x in queries:
            res = [0] * k
            res[v % k] += 1
            sg.set(i, (v % k, res))
            _, cnt = sg.prod(s, n)
            ans.append(cnt[x])
        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])
LeetCode Weekly Contest 447
LeetCode Weekly Contest 445