feat: Added Weighted Interval Scheduling

DIRECTORY.md

@@ -1195,6 +1195,7 @@
   * [Non Preemptive Shortest Job First](scheduling/non_preemptive_shortest_job_first.py)
   * [Round Robin](scheduling/round_robin.py)
   * [Shortest Job First](scheduling/shortest_job_first.py)
+  * [Weighted Interval Scheduling](scheduling/weighted_interval_scheduling.py)
 
 ## Searches
   * [Binary Search](searches/binary_search.py)

scheduling/weighted_interval_scheduling.py

@@ -0,0 +1,81 @@
+# 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.
+# The goal is to find the maximum weight subset of non-overlapping jobs.
+# 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.
+        Return: The index of the latest non-conflicting job.
+    >>> latest_non_conflict([(1, 3, 50), (2, 5, 20), (4, 6, 30)], 2)
+    0
+    >>> latest_non_conflict([(1, 3, 50), (3, 4, 60), (5, 9, 70)], 2)
+    1
+    """
+    for j in range(n - 1, -1, -1):
+        if jobs[j][1] <= jobs[n][0]:
+            return j
+    return -1
+
+
+def find_max_weight(jobs: list[tuple[int, int, int]]) -> int:
+    """
+    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
+    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)])
+    80
+    >>> find_max_weight([(1, 3, 10), (2, 5, 100), (6, 8, 15)])
+    115
+    >>> find_max_weight([(1, 3, 20), (3, 5, 30), (6, 19, 60), (2, 100, 200)])
+    200
+    """
+    # Sort jobs based on their end times
+    jobs.sort(key=lambda x: x[1])
+
+    # Initialize dp array to store the maximum weight up to each job
+    n = len(jobs)
+    dp = [0] * n
+    dp[0] = jobs[0][2]  # The weight of the first job is the initial value
+
+    for i in range(1, n):
+        # Include the current job
+        include_weight = jobs[i][2]
+        latest_job = latest_non_conflict(jobs, i)
+        if latest_job != -1:
+            include_weight += dp[latest_job]
+
+        # 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
+
+
+if __name__ == "__main__":
+    # Example list of jobs (start_time, end_time, weight)
+    jobs = [(1, 2, 50), (3, 5, 20), (6, 19, 100), (2, 100, 200)]
+
+    # Ensure we have jobs to process
+    if len(jobs) == 0:
+        print("No jobs available to process")
+        raise SystemExit(0)
+
+    # Calculate the maximum weight for non-overlapping jobs
+    max_weight = find_max_weight(jobs)
+
+    # Print the result
+    print(f"The maximum weight of non-overlapping jobs is {max_weight}")