Possibility Of Finishing
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.graphs;
import java.util.*;
/*
* ============================================================================================
* PROBLEM: Possibility of Finishing (Course Schedule)
* ============================================================================================
*
* ---------------------
* 1. Problem Description
* ---------------------
* There are 'A' courses labeled 1 to A. You are given prerequisites in the form
* of pairs (B[i], C[i]), meaning you must take course B[i] before C[i].
* Determine if it is possible to finish all courses.
*
* ---------------------
* 2. Logical Breakdown & Mathematical Formulation
* ---------------------
* This problem translates to: "Does the directed graph contain a cycle?"
* - If there is a cycle (e.g., 1 -> 2 -> 3 -> 1), you can never start any course
* in that cycle because each depends on another.
* - If the graph is a Directed Acyclic Graph (DAG), it is always possible to
* finish the courses.
*
* Algorithm (Kahn's Algorithm / BFS):
* 1. Build an adjacency list and calculate the in-degree of every node.
* 2. Add all nodes with in-degree 0 to a Queue (courses with no prerequisites).
* 3. While the Queue is not empty:
* a. Pop course 'u'.
* b. Increment a counter count.
* c. For each neighbor 'v' of 'u' (courses that depend on 'u'):
* - Decrement in-degree of 'v'.
* - If in-degree of 'v' becomes 0, add 'v' to the Queue.
* 4. If count == A, return 1 (Possible). Otherwise, return 0 (Cycle detected).
*
* ---------------------
* 3. Constraints & Complexity Analysis
* ---------------------
* Constraints:
* 1 <= A <= 10^5
* 1 <= length(B), length(C) <= 10^5
*
* Time Complexity:
* O(A + M) - Where M is the number of prerequisite pairs. We visit every node
* and edge once.
*
* Space Complexity:
* O(A + M) - To store the adjacency list and the in-degree array.
*
* ============================================================================================
/
public class PossibilityOfFinishing {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
public int solve(int i, int[] b5, int[] c5) {
return 1;
}
}