Reversing Edges
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;
/* * ============================================================================================ * PROBLEM: Reversing Edges * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given a directed graph with A nodes and M edges. * Find the minimum number of edges that need to be reversed to create a path * from Node 1 to Node A. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * We can model this problem as finding the shortest path in a weighted graph. * * Step 1: Graph Transformation * For every directed edge u -> v in the original graph: * - Create a directed edge u -> v with weight 0. (Cost of using existing edge is 0). * - Create a directed edge v -> u with weight 1. (Cost of reversing this edge is 1). * * Step 2: Shortest Path Calculation * - Use Dijkstra's Algorithm (or 0-1 BFS since weights are only 0 and 1) starting * from Node 1. * - The shortest distance to Node A represents the minimum number of "weight 1" * edges (reversals) required. * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= A, M <= 10^5 * * Time Complexity: * O((A + M) log A) - Using Dijkstra's with a Priority Queue. * O(A + M) - If using 0-1 BFS with a Deque (Double-ended queue). * * Space Complexity: * O(A + M) - To store the expanded adjacency list and distance array. * * --------------------- * 4. Example * --------------------- * Edge: 1 -> 2, 3 -> 2 * Transformed: * (1, 2) weight 0 * (2, 1) weight 1 * (3, 2) weight 0 * (2, 3) weight 1 * Path 1 -> 2 -> 3: Total weight = 0 + 1 = 1 reversal. * ============================================================================================ / public class ReversingEdges { public int solve(int A, int[][] edges3) { // TODO: Implement solution return 0; } }