Chapter 6: Sorting and Searching

# 知识解析

6.4 Binary Search

这里参照 Ruby 中 bsearch_index 的实现，take一个 function 作为参数表示需要满足的条件，找到满足条件的最小index和最大index（注意这一条件对 bsearch_low 必须是“单调递增“，对 bsearch_high 必须是“单调递减“）。

如果寻找的是具体的一个target值，需要传入 lambda ele: ele >= target 或 lambda ele: ele <= target 在找到index的上下限之后需要额外一步检查来确定是否能找到。

"""
condition should be a function that evaluates as
[False, False ... True, True, True], aka >= target
"""


def bsearch_low(arr: List[int], condition: Callable[[int], bool]) -> Optional[int]:
    low = 0
    high = len(arr) - 1
    while low <= high:
        mid = (low + high) // 2  
        smaller = condition(arr[mid])
        if smaller:
            high = mid - 1
        else:
            low = mid + 1

    if low >= len(arr) or not condition(arr[low]):
        return None

    return low


"""
condition should be a function that evaluates as
[True, True ... False, False, False], aka <= target
"""


def bsearch_high(arr: List[int], condition: Callable[[int], bool]) -> Optional[int]:
    low = 0
    high = len(arr) - 1
    while low <= high:
        mid = (low + high) // 2  
        bigger = condition(arr[mid])
        if bigger:
            low = mid + 1
        else:
            high = mid - 1

    if high < 0 or not condition(arr[high]):
        return None

    return high

e.g. 1 Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. ( LeetCode: Find First and Last Position of Element in Sorted Array )

from bsearch import bsearch_low, bsearch_high
from typing import List


def search_range(nums: List[int], target: int) -> List[int]:
    lower_bound = bsearch_low(nums, lambda ele: ele >= target)
    if lower_bound is None or (nums[lower_bound] != target):
        return [-1, -1]

    return [lower_bound, bsearch_high(nums, lambda ele: ele <= target)]

e.g.2 We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i]. You're given the startTime , endTime and profit arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.( LeetCode: Maximum Profit in Job Scheduling)

from bsearch import bsearch_low, bsearch_high

def jobScheduling(startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        return MaxProfitFinder(startTime, endTime, profit).max_profit()


class Job:
    def __init__(self, start_time, end_time, profit):
        self.start_time = start_time
        self.end_time = end_time
        self.profit = profit


class MaxProfitFinder:
    def __init__(self, start_times, end_times, profits):
        self.jobs = [
            Job(start_times[idx], end_times[idx], profits[idx])
            for idx in range(len(start_times))
        ]
        self.jobs.sort(key=lambda job: job.start_time)

    def find_next_job_id(self, time):
        return bsearch_low(self.jobs, lambda job: job.start_time >= time)

    def max_profit(self):
        return self.max_profit_from_id(0)

    @lru_cache(maxsize=None)
    def max_profit_from_id(self, id):
        if id is None or id >= len(self.jobs):
            return 0

        job = self.jobs[id]
        next_job_id = self.find_next_job_id(job.end_time)
        max_profit = max(
            job.profit + self.max_profit_from_id(next_job_id),
            self.max_profit_from_id(id + 1)
        )

        return max_profit