Another Coin Problem
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.heaps;
import java.util.*;
/*
* ============================================================================================
* PROBLEM: Another Coin Problem (Greedy Change Making)
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* We need to represent the integer A as a sum of powers of 5:
* $\(A = c_k \cdot 5^k + c_{k-1} \cdot 5^{k-1} + \dots + c_1 \cdot 5^1 + c_0 \cdot 5^0\)$
* To minimize \(\sum c_i\), we should use the largest denominations first.
*
* Step 1: Base-5 Conversion Analogy
* - This problem is essentially asking for the sum of digits of A when expressed
* in Base-5.
* - For example, if A = 31, in Base-5 it is \((111)_5\) because \(1(25) + 1(5) + 1(1) = 31\).
* - Number of coins = \(1 + 1 + 1 = 3\).
*
* Step 2: Algorithm
* 1. While A > 0:
* 2. Find the largest power of 5 (let's say \(5^k\)) that is \(\le A\).
* 3. Use as many coins of \(5^k\) as possible: count += A / 5^k.
* 4. Update the remaining amount: A %= 5^k.
* 5. Move to the next lower power of 5.
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O(logâ‚… A)
* - The number of iterations is proportional to the number of digits of A in base 5.
*
* Space Complexity:
* O(1)
* - We only need a few variables to track the current power and the total count.
*
* ---------------------
* 3. Base Case Verification
* ---------------------
* - A = 1: Returns 1 (one \(5^0\) coin).
* - A = 5: Returns 1 (one \(5^1\) coin).
* - A = 0: Returns 0.
* ============================================================================================
/
public class AnotherCoinProblem {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
}