Cows And Snacks
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: Cows and Snacks * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * There are A flavors of snacks (1 to A) and M guests. Each guest has exactly two favorite * flavors. Guests take turns eating all remaining snacks of their favorite flavors. * If no snacks of their favorite flavors are left, the guest becomes sad. * * Task: * Minimize the number of sad guests by ordering them optimally. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * We can model this using a Graph: * - Nodes: The A snack flavors. * - Edges: Each guest represents an edge connecting their two favorite flavors (u, v). * * Key Insight: * Consider a connected component with 'V' nodes (flavors) and 'E' edges (guests). * 1. To satisfy a guest, we must use at least one new snack flavor. * 2. In any connected component with V nodes, we can satisfy at most V - 1 guests. * - This is because the first guest eats 2 snacks, and every subsequent guest * in the optimal order only needs 1 snack to be satisfied. * - To satisfy V nodes, we need a "Spanning Tree" structure, which uses V-1 edges. * * Mathematical Formula: * - Let 'K' be the number of connected components in the graph. * - Total satisfied guests = (Total Nodes in component_1 - 1) + (Total Nodes in component_2 - 1) + ... * - Total satisfied guests = (Total Nodes) - (Number of Components) = A - K. * * Number of Sad Guests: * Total Guests (M) - Total Satisfied Guests (A - K). * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= A, M <= 10^5 * * Time Complexity: * O(A + M) - Using Disjoint Set Union (DSU) with path compression or DFS to find components. * * Space Complexity: * O(A + M) - To store the adjacency list or DSU parent array. * * ============================================================================================ / public class CowsAndSnacks { public int solve(int A, int[][] b4) { // TODO: Implement solution return 0; } }