Tushars Birthday Party

Logical Breakdown

• [ ] Subproblem Identification:
• [ ] Optimal Substructure:
• [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
• [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

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Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

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

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Visualization

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

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Complexity Analysis

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

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Code Reference

package com.dsa.dynamic_programming;

import java.util.*;

/ * Problem: Tushars Birthday Party * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Tushar's Birthday Party * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Input: * - A: Array of integers (Capacity needed for each friend). * - B: Array of integers (Filling capacity of each dish). * - C: Array of integers (Cost of each dish). * * Goal: Minimize total cost to satisfy ALL friends. * Note: If a friend needs capacity K, the sum of filling capacities of dishes eaten * must be exactly (or at least) K. (Problem usually implies exact or 'reaching' state). * Since this is "Unbounded Knapsack" logic, usually we look for exact fill or best fill. * In this specific problem context, it's typically "Min Cost Unbounded Knapsack to reach Weight W". * * --------------------- * 2. Logical Breakdown * --------------------- * This is "Min Cost Unbounded Knapsack". * Instead of Maximizing Value for Weight <= W, we Minimize Cost for Weight == W. * * Algorithm: * 1. Find the Maximum capacity required among all friends (MaxCap). * 2. Solve the DP for all weights from 1 to MaxCap. * - dp[w] = Min cost to reach filling capacity 'w'. * - Initialize dp[0] = 0, others = Infinity. * - Recurrence: dp[w] = min(dp[w], cost[i] + dp[w - filling[i]]) * 3. After filling DP table, sum up dp[A[k]] for all friends k. * * --------------------- * 3. Complexity * --------------------- * Time: O(N_dishes * Max_Friend_Capacity) * Space: O(Max_Friend_Capacity) * * ============================================================================================ */ public class TusharsBirthdayParty { public int solve(int A) { // TODO: Implement solution return 0; }

public int solve(int[] friends, int[] filling, int[] cost) {
    return 1;
}

}