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B Karthik 2024-10-05 08:35:44 +05:30
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scheduling/weighted_interval_scheduling.py
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@ -1,16 +1,20 @@
# Implementation of Weighted Interval Scheduling algorithm
# In this algorithm, we are given a list of jobs with start and end times, and each job has a specific weight.
# In this algorithm, we are given a list of jobs with start and end times,
# and each job has a specific weight.
# The goal is to find the maximum weight subset of non-overlapping jobs.
# https://en.wikipedia.org/wiki/Interval_scheduling#:~:text=their%20finishing%20times.-,Weighted,-%5Bedit%5D
# https://en.wikipedia.org/wiki/Interval_scheduling
from __future__ import annotations
def latest_non_conflict(jobs: list[tuple[int, int, int]], n: int) -> int:
"""
This function finds the latest job that does not conflict with the current job at index `n`.
The jobs are given as (start_time, end_time, weight), and the jobs should be sorted by end time.
It returns the index of the latest job that finishes before the current job starts.
This function finds the latest job that does not conflict with
the current job at index `n`.
The jobs are given as (start_time, end_time, weight), and the
jobs should be sorted by end time.
It returns the index of the latest job that finishes before the
current job starts.
Return: The index of the latest non-conflicting job.
>>> latest_non_conflict([(1, 3, 50), (2, 5, 20), (4, 6, 30)], 2)
0
@ -25,9 +29,11 @@ def latest_non_conflict(jobs: list[tuple[int, int, int]], n: int) -> int:
def find_max_weight(jobs: list[tuple[int, int, int]]) -> int:
"""
This function calculates the maximum weight of non-overlapping jobs using dynamic programming.
This function calculates the maximum weight of non-overlapping jobs
using dynamic programming.
Each job is represented by a tuple (start_time, end_time, weight).
The function builds a DP table where each entry `dp[i]` represents the maximum weight achievable
The function builds a DP table where each entry `dp[i]` represents
the maximum weight achievable
using jobs from index 0 to i.
Return: The maximum achievable weight without overlapping jobs.
>>> find_max_weight([(1, 3, 50), (2, 5, 20), (4, 6, 30)])
@ -52,7 +58,8 @@ def find_max_weight(jobs: list[tuple[int, int, int]]) -> int:
if latest_job != -1:
include_weight += dp[latest_job]
# Exclude the current job, and take the maximum of including or excluding
# Exclude the current job, and take the maximum of including or
# excluding
dp[i] = max(include_weight, dp[i - 1])
return dp[-1] # The last entry contains the maximum weight