Max Wine Profit
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.dynamic_programming;
import java.util.*;
/ * Problem: Max Wine Profit * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Max Wine Profit * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given N wines in a row. Each year you can sell one bottle from either the LEFT or RIGHT end. * Prices increase over time: Year 1 multiplier is 1, Year 2 is 2, etc. * Price = wine_cost * current_year. * Find max profit. * * --------------------- * 2. Logical Breakdown * --------------------- * Let dp[L][R] be the max profit obtainable from the subarray wines[L...R]. * * Calculation of Year: * Total bottles = N. * Current bottles remaining = (R - L + 1). * Current Year = N - (R - L + 1) + 1. * * Recurrence: * 1. Sell Left (wine[L]): * Profit = wine[L]Year + dp[L+1][R] * 2. Sell Right (wine[R]): * Profit = wine[R]Year + dp[L][R-1] * * dp[L][R] = max(Sell Left, Sell Right) * * Base Case: * L == R (Last bottle): Profit = wine[L] * N. * * --------------------- * 3. Complexity * --------------------- * Time: O(N^2) * Space: O(N^2) * * ============================================================================================ */ public class MaxWineProfit { public int solve(int A) { // TODO: Implement solution return 0; }
public int solveWine(int[] wines) {
return 1;
}
}