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Construction Cost

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

\[ dp[i] = ... \]

Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

Complexity Analysis

Approach Time Complexity Space Complexity
Iterative DP \(O(N)\) \(O(1)\)

Code Reference

package com.dsa.graphs;

/* * ============================================================================================ * PROBLEM: Construction Cost (Minimum Spanning Tree) * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given 'A' distribution centers (nodes) and 'C' possible roads (weighted edges). * Find the minimum cost to connect all centers such that every center is reachable * from the first one. This is equivalent to finding the Minimum Spanning Tree (MST). * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * We implement Kruskal's Algorithm using Disjoint Set Union (DSU). * * Step 1: Sorting * Sort all edges in ascending order of their weights. Since we want minimum cost, * we always prioritize the cheapest available edge. * * Step 2: DSU with Path Compression and Union by Rank * To efficiently detect cycles and merge components: * - find(i): Returns the root of the component containing node i. * - union(i, j): Merges components containing i and j. * * Step 3: Kruskal's Execution * 1. Iterate through sorted edges. * 2. For an edge (u, v) with weight w: * - If find(u) != find(v): * - These nodes are in different components. * - Adding this edge will not form a cycle. * - Add w to totalCost. * - Perform union(u, v). * 3. Continue until all nodes are connected (edges added = A-1). * * Step 4: Modulo Arithmetic * As edge weights can be up to 10^9 and A up to 10^5, the total cost can exceed * the capacity of a 64-bit long. Perform all additions modulo 10^9 + 7. * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Time Complexity: O(C log C + C * α(A)) * - Sorting: O(C log C) * - DSU Operations: O(C * α(A)), where α is the inverse Ackermann function. * * Space Complexity: O(A + C) * - Adjacency/Edge list: O(C) * - DSU parent and rank arrays: O(A) * * ============================================================================================ / public class ConstructionCost { public int solve(int A, int[][] b3) { // TODO: Implement solution return 0; } }