feat: Add Weighted Job Scheduling algorithm to dynamic programming

DIRECTORY.md

@@ -26,6 +26,7 @@
   * [Word Break](backtracking/word_break.py)
   * [Word Ladder](backtracking/word_ladder.py)
   * [Word Search](backtracking/word_search.py)
+  * [Weighted Job Scheduling](backtracking/weighted_job_scheduling.py)
 
 ## Bit Manipulation
   * [Binary And Operator](bit_manipulation/binary_and_operator.py)
@@ -880,6 +881,8 @@
   * [Sdes](other/sdes.py)
   * [Tower Of Hanoi](other/tower_of_hanoi.py)
   * [Word Search](other/word_search.py)
+  * [Weighted Job Scheduling](backtracking/weighted_job_scheduling.py)
+
 
 ## Physics
   * [Altitude Pressure](physics/altitude_pressure.py)

dynamic_programming/weighted_job_scheduling.py

@@ -0,0 +1,77 @@
+"""
+Author  : Prince Kumar Prajapati
+Date    : October 9, 2025
+
+Weighted Job Scheduling Problem
+--------------------------------
+Given N jobs where every job has a start time, finish time, and profit,
+find the maximum profit subset of jobs such that no two jobs overlap.
+
+Approach:
+- Sort all jobs by their finish time.
+- For each job, find the last non-conflicting job using binary search.
+- Use Dynamic Programming to build up the maximum profit table.
+
+Time Complexity:  O(n log n)
+Space Complexity: O(n)
+"""
+
+
+def find_last_non_conflicting(jobs, index):
+    """
+    Binary search to find the last job that doesn't overlap
+    with the current job (at index).
+    """
+    low, high = 0, index - 1
+    while low <= high:
+        mid = (low + high) // 2
+        if jobs[mid][1] <= jobs[index][0]:
+            if mid + 1 < index and jobs[mid + 1][1] <= jobs[index][0]:
+                low = mid + 1
+            else:
+                return mid
+        else:
+            high = mid - 1
+    return -1
+
+
+def weighted_job_scheduling(jobs):
+    """
+    Function to find the maximum profit from a set of jobs.
+
+    Parameters:
+        jobs (list of tuples): Each tuple represents (start_time, end_time, profit)
+    Returns:
+        int: Maximum profit achievable without overlapping jobs.
+    """
+
+    # Step 1: Sort jobs by their finish time
+    jobs.sort(key=lambda x: x[1])
+
+    n = len(jobs)
+    dp = [0] * n  # dp[i] will store the max profit up to job i
+    dp[0] = jobs[0][2]  # First job profit is the base case
+
+    # Step 2: Iterate through all jobs
+    for i in range(1, n):
+        # Include current job
+        include_profit = jobs[i][2]
+
+        # Find the last non-conflicting job
+        j = find_last_non_conflicting(jobs, i)
+        if j != -1:
+            include_profit += dp[j]
+
+        # Exclude current job (take previous best)
+        dp[i] = max(include_profit, dp[i - 1])
+
+    # Step 3: Return the maximum profit at the end
+    return dp[-1]
+
+
+# Example usage / Test case
+if __name__ == "__main__":
+    # Each job is represented as (start_time, end_time, profit)
+    jobs = [(1, 3, 50), (2, 4, 10), (3, 5, 40), (3, 6, 70)]
+
+    print("Maximum Profit:", weighted_job_scheduling(jobs))