# Money change problem: Greedy vs. Dyn.Pro.

This is a classical problem of Computer Science: it’s used to study both Greedy and Dynamic Programming algorithmic techniques.

## Definition

Given:

- A set of
`n`

Denominations`D[0...n-1]`

in ascending order, representing a Monetary Coin System - An money amount
`A`

, as input

calculate a solution:

, with`S[0...n-1]`

`0 <= S[i] <= (A/S[i])`

and`0 < i < n-1`

where:

`A = Sum`

_{[i=0 -> n-1]}{ D[i] * S[i] }`Min{ Sum`

_{[i=0 -> n-1]}{ S[i] } }

## In other words

Find the smallest amount of coins to make the given amount.

## First, the Greedy solution

The Greedy approach is as expected: tries to take as much largest coins as possible. Nothing fancy. [soucecode:c] change_coins_greedy(D[], A): init S[n] i = n-1 // Pick as much largest coins as possible while ( A > 0 ) do: S[i] = A / D[i] A = A - S[i] * D[i] i = i - 1 endwhile

// Set to ‘0’ the result for all the other coins while ( i >= 0 ) do: S[i] = 0 i = i - 1 endwhile

```
This algorithm, of time complexity <strong>O(A)</strong>, doesn't work for some (rare) situations.
## When Greedy is not enough
<del datetime="2010-01-18T10:41:49+00:00">The Greedy algorithm doesn't work for Denominations where if <strong>2 denominations D[i] and D[j] exists</strong> with:</del>
<ul>
<li><del datetime="2010-01-18T10:41:49+00:00"><strong>i < j</strong></del></li>
<li><del datetime="2010-01-18T10:41:49+00:00"><strong>D[i] < D[j]</strong></del></li>
<li><del datetime="2010-01-18T10:41:49+00:00"><strong>2 * D[i] > D[j]</strong></del></li>
</ul>
<del datetime="2010-01-18T10:41:49+00:00">For example, taken <strong>D = {1, 10, 30, 40}</strong> and amount <strong>A = 63</strong>, the Greedy algorithm will build a solution <strong>S = {3, 2, 0, 1}</strong>, that is sub-optimal. The optimal solution in this case is <strong>S = {3, 0, 2, 0}</strong>.</del>
<strong>UPDATE:</strong> <a href="#comment-11505">In the comments Vincenzo</a> gives an example where this condition doesn't still stand but the Greedy Algorithm still produces the best solution.
## Dynamically Programmed Solution
In real life the Greedy algorithm should be always enough: I couldn't find any money system that has the problem described above. And, indeed, the Greedy approach is what every human being "normally" applies when changing money.
But we are Comp-Sci, and we need to find a better solution ;-) - a Dynamically Programmed one.
Given the function <strong>M[j], that is the minimum number of coins to make the amount 'j'</strong>, it looks like:
<ul>
<li><strong>M[A] = min<sub>[i = 0 -> n-1]</sub> { M[ A - D[i] ] +1 , M[A] }</strong></li>
</ul>
Here is the code:
```c
unsigned int* change_coins_dynpro(unsigned short int D[],
unsigned int D_size, unsigned int amount) {
// Min. num. of coins for the given 'amount'
unsigned int min_num_coins[amount], cur_min_num_coins;
// Biggest coin used for the given 'amount'
unsigned int biggest_coin_at[amount], cur_biggest_coin_at;
unsigned int i, j;
unsigned int* solution = NULL;
// Ensure the Denomination System can represent any value
if ( D[0] != 1 ) return NULL;
// Initialize the solution array to Zero
solution = (unsigned int*)calloc(D_size, sizeof(unsigned int));
if ( NULL == solution ) {
return NULL;
}
// Amount '0' requires '0' coins
min_num_coins[0] = 0;
biggest_coin_at[0] = 0;
// For Amounts from '1' to 'amount'
for ( i = 1; i <= amount; ++i ) {
// Start taking 'D[0]' 'i-times'
cur_min_num_coins = (i / D[0]);
cur_biggest_coin_at = 0;
// For coins from 'D[1]' to 'D[D_size -1]'
for ( j = 1; j < D_size; ++j ) {
// If 'D[j]' minimizes the num. of coins to take for amount 'i'
if ( (i >= D[j]) && (cur_min_num_coins >= (min_num_coins[ i - D[j] ] +1)) ) {
cur_min_num_coins = min_num_coins[ i - D[j] ] +1;
cur_biggest_coin_at = j;
}
}
// Store the minimum just found
min_num_coins[i] = cur_min_num_coins;
biggest_coin_at[i] = cur_biggest_coin_at;
}
// Let's build the solution array
while ( amount > 0 ) {
cur_biggest_coin_at = biggest_coin_at[amount];
solution[ cur_biggest_coin_at ] += 1; // Add '1' of this coin to the solution
amount -= D[cur_biggest_coin_at]; // Amount left when picking this coin
}
return solution;
}
```

Time complexity for this algorithm is **O( Amount * num_of_denominations )**.
*Pretty heavy algorithm, but do you want to compare with the satisfaction of carrying the minimum amount of coins with you? :P*