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Dice Throw

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.dynamic_programming;

import java.util.*;

/ * Problem: Dice Throw * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Dice Throw (Sum A) * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Find the number of ways to obtain a specific Sum A by throwing a standard dice (1-6) * repeatedly. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Let dp[i] be the number of ways to get a sum of exactly 'i'. * * To reach sum 'i', the last throw could have been 1, 2, 3, 4, 5, or 6. * - If last throw was 1, we needed sum (i-1) previously. * - If last throw was 2, we needed sum (i-2) previously. * ... * * Recurrence Relation: * dp[i] = dp[i-1] + dp[i-2] + dp[i-3] + dp[i-4] + dp[i-5] + dp[i-6] * (For k from 1 to 6, add dp[i-k] if i-k >= 0). * * Base Case: * dp[0] = 1 (0 ways to throw dice, but sum 0 is the starting state). * * --------------------- * 3. Constraints & Complexity * --------------------- * Time Complexity: O(A) (Specifically 6A). * Space Complexity: O(A). * * ============================================================================================ / public class DiceThrow { public int solve(int A) { // TODO: Implement solution return 0; }

public int solveDice(int i) {
    return 0;
}

}