Gym Trainer
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: Gym Trainer * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * There are A people in a gym. Friends are formed through two distinct activities: * 1. Walking together (represented by edge set B). * 2. Talking together (represented by edge set C). * * Key Constraint: * A person cannot walk with one friend and talk with another. This implies that if a person * belongs to a "walking group", they cannot be part of a "talking group" and vice versa. * However, they can walk with multiple people or talk with multiple people. * * Task: * Suggest one of 2 possible diets to each person. All friends in a connected component * must have the same diet. Return total ways modulo 10^9 + 7. * If the constraint is violated, return 0. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Step 1: Constraint Validation * - Use two separate structures (or Disjoint Set Union) to track "walking" and "talking" groups. * - For every person i from 1 to A: * - Check if person i has at least one "walking" neighbor AND at least one "talking" neighbor. * - If yes, the possessive constraint is violated. Return 0. * * Step 2: Component Counting * - Merge all friends (both walking and talking) into a single graph. * - Even though a person can't do both activities, friends of friends must still * share the same diet. * - Let 'K' be the total number of connected components in the combined graph. * * Step 3: Calculation * - Each component can choose one of 2 diets independently. * - Total Ways = 2^K (modulo 10^9 + 7). * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= A <= 10^5 * 1 <= |B|, |C| <= 10^5 * * Time Complexity: * O(A + |B| + |C|) - Using DSU with path compression or DFS for component counting. * * Space Complexity: * O(A + |B| + |C|) - To store the graph/DSU parent array. * * ============================================================================================ / public class GymTrainer { public int solve(int A) { // TODO: Implement solution return 0; }
public int solve(int i, int[][] b4, int[][] c4) {
return 1;
}
}