Construct Roads
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: Construct Roads * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * A country has N cities connected by N-1 roads. Since N cities are connected by N-1 roads * and all cities are reachable, the initial structure is a Tree. * * Goal: Add the maximum number of roads such that the resulting graph remains Bipartite. * * A Bipartite graph is one where vertices can be divided into two disjoint sets U and V * such that every edge connects a vertex in U to one in V. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * 1. Property of Trees: * Every tree is inherently a Bipartite graph. We can partition the nodes into two * sets (Set1 and Set2) using a simple 2-coloring algorithm (BFS or DFS). * * 2. Maximum Edges in a Bipartite Graph: * In a complete bipartite graph with sets of size n1 and n2, the maximum number * of edges possible is (n1 * n2). * * 3. Calculation: * - Let n1 = count of nodes in Set1. * - Let n2 = count of nodes in Set2. * - Max possible edges = (n1 * n2). * - Existing edges = N - 1. * - New roads to construct = (n1 * n2) - (N - 1). * * 4. Modulo Arithmetic: * Since n1 * n2 can exceed 10^9, use long for calculations and return result % (10^9 + 7). * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= N <= 10^5 * * Time Complexity: * O(N) - We perform a single BFS/DFS traversal to color the nodes and count set sizes. * * Space Complexity: * O(N) - To store the adjacency list and the color/visited array. * * --------------------- * 4. Example Walkthrough * --------------------- * Input: A = 5, B = [[1, 3], [1, 4], [3, 2], [3, 5]] * - Color 1: Red (Set1) * - Neighbors of 1 (3, 4): Blue (Set2) * - Neighbors of 3 (2, 5): Red (Set1) * Set1: {1, 2, 5} -> n1 = 3 * Set2: {3, 4} -> n2 = 2 * Max Edges = 3 * 2 = 6. * Existing Edges = 4. * New Roads = 6 - 4 = 2. * ============================================================================================ / public class ConstructRoads { public int solve(int A, int[][] edges4) { // TODO: Implement solution return 0; } }