LCP 08. 剧情触发时间

合集 - LeetCode 二分法(17)

1.35. 搜索插入位置2024-09-092.704. 二分查找2024-09-093.744. 寻找比目标字母大的最小字母2024-09-094.2529. 正整数和负整数的最大计数2024-09-095.1385. 两个数组间的距离值2024-09-096.2300. 咒语和药水的成功对数2024-09-097.2389. 和有限的最长子序列2024-09-098.1170. 比较字符串最小字母出现频次2024-09-099.2080. 区间内查询数字的频率2024-09-0910.2563. 统计公平数对的数目2024-09-0911.981. 基于时间的键值存储2024-09-1012.1146. 快照数组2024-09-1013.658. 找到 K 个最接近的元素2024-09-1014.1818. 绝对差值和2024-09-1015.911. 在线选举2024-09-10

16.LCP 08. 剧情触发时间2024-09-10

17.[模板题] - 300. 最长递增子序列2024-09-13

收起

题目链接	LCP 08. 剧情触发时间
思路	前缀和+二分法
题解链接	python 前缀和+二分法
关键点	预处理：前处理得到各个时刻三种资源的累计值（必为升序数组）；查找：二分法查找三种资源需要满足的时刻，取三者最大值即可得到答案
时间复杂度	$O(n)$
空间复杂度	$O(n)$

代码实现：

class Solution:
    def getTriggerTime(self, increase: List[List[int]], requirements: List[List[int]]) -> List[int]:
        n = len(increase)
        presum = [
            [0] * (n+1) for _ in range(3)
        ]
        for i, (c, r, h) in enumerate(increase):
            presum[0][i+1] = presum[0][i] + c
            presum[1][i+1] = presum[1][i] + r
            presum[2][i+1] = presum[2][i] + h
        
        # found the left-most index `nums[index] >= val`
        def lower_bound(nums, val):
            left, right = -1, n+1
            while left + 1 < right:
                mid = (left+right) // 2
                if nums[mid] < val:
                    left = mid
                else:
                    right = mid
            return right
 
        m = len(requirements)
        answer = [-1] * m
        for i, (c, r, h) in enumerate(requirements):
            x = lower_bound(presum[0], c)
            y = lower_bound(presum[1], r)
            z = lower_bound(presum[2], h)
            min_index = max(x, y, z)
            if min_index <= n:
                answer[i] = min_index
        return answer

Python-标准库

class Solution:
    def getTriggerTime(self, increase: List[List[int]], requirements: List[List[int]]) -> List[int]:
        n = len(increase)
        presum = [
            [0] * (n+1) for _ in range(3)
        ]
        for i, (c, r, h) in enumerate(increase):
            presum[0][i+1] = presum[0][i] + c
            presum[1][i+1] = presum[1][i] + r
            presum[2][i+1] = presum[2][i] + h
        
        m = len(requirements)
        answer = [-1] * m
        for i, (c, r, h) in enumerate(requirements):
            x = bisect_left(presum[0], c)
            y = bisect_left(presum[1], r)
            z = bisect_left(presum[2], h)
            min_index = max(x, y, z)
            if min_index <= n:
                answer[i] = min_index
        return answer