Massive Algorithms: Dynamic Programming | Set 7 (Coin Change)

Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter.
From http://www.algorithmist.com/index.php/Coin_Change
To count total number solutions, we can divide all set solutions in two sets.
1) Solutions that do not contain mth coin (or Sm).
2) Solutions that contain at least one Sm.
Let count(S[], m, n) be the function to count the number of solutions, then it can be written as sum of count(S[], m-1, n) and count(S[], m, n-Sm).

Recursive Formulation

We are trying to count the number of distinct sets.

Since order does not matter, we will impose that our solutions (sets) are all sorted in non-decreasing order (Thus, we are looking at sorted-set solutions: collections).

For a particular

N

and $S = \{ S_1, S_2, \ldots, S_m \}$ (now with the restriction that $S_1 < S_2 < \ldots < S_m$ , our solutions can be constructed in non-decreasing order), the set of solutions for this problem,

C (N, m)

, can be partitioned into two sets:

There are those sets that do not contain any $S m$ and
Those sets that contain at least 1 $S m$

This partitioning will essentially break the initial problem into two subproblems:

If $N < S m$ (that is, a solution does not contain $S m$ ), then we can solve the subproblem of $N$ with $S = \{ S_1, S_2, \ldots, S_{m-1} \}$ , or the solutions of $C (N, m - 1)$ .
If $N \geq S_m$ (that is, a solution does in fact contain $S m$ ), then we are using at least one $S m$ , thus we are now solving the subproblem of $N - S m$ , $S = \{ S_1, S_2, \ldots, S_m \}$ . This is $C (N - S m, m)$ .

Thus, we can formulate the following:

C (N, m) = C (N, m - 1) + C (N - S m, m)

Time Complexity: O(mn)

int count( int S[], int m, int n )

{

    int i, j, x, y;

    // We need n+1 rows as the table is consturcted in bottom up manner using 

    // the base case 0 value case (n = 0)

    int table[n+1][m];

    // Fill the enteries for 0 value case (n = 0)

    for (i=0; i<m; i++)

        table[0][i] = 1;

    // Fill rest of the table enteries in bottom up manner  

    for (i = 1; i < n+1; i++)

    {

        for (j = 0; j < m; j++)

        {

            // Count of solutions including S[j]

            x = (i-S[j] >= 0)? table[i - S[j]][j]: 0;

            // Count of solutions excluding S[j]

            y = (j >= 1)? table[i][j-1]: 0;

            // total count

            table[i][j] = x + y;

        }

    }

    return table[n][m-1];

}

Following is a simplified version of method 2. The auxiliary space required here is O(n) only.

http://stackoverflow.com/questions/27880842/space-optimized-solution-for-coin-change
It just uses previous row, same column value, and same row, previous columns(this is always kept).

int count( int S[], int m, int n )

{

    // table[i] will be storing the number of solutions for

    // value i. We need n+1 rows as the table is consturcted

    // in bottom up manner using the base case (n = 0)

    int table[n+1];

    // Initialize all table values as 0

    memset(table, 0, sizeof(table));

    // Base case (If given value is 0)

    table[0] = 1;

    // Pick all coins one by one and update the table[] values

    // after the index greater than or equal to the value of the

    // picked coin

    for(int i=0; i<m; i++)

        for(int j=S[i]; j<=n; j++)

            table[j] += table[j-S[i]];

    return table[n];

}

http://www.zrzahid.com/coin-sum-problem-dynamic-programming/

Optimal Substructure
We can see that we can find the number ways we can make n using C={ c1, c2, .. , cm} coins by combining solution of making (n-cm) and including last coin cm to current solution; or we can simply ignore the last coin cm and try to find solution with remaining coins.

let, cs(C, n, m) be the number of solutions of making n using m coins C = {c1, c2, …, cm}. The we can write the following recursive formula,

cs(C, n, m) = cs(C, n-cm, m) + cs(C, n, m-1).

public static int coinSum(final int[] coins, final int sum) {
    final int m = coins.length;
    final int[][] csTable = new int[sum + 1][m + 1];

    // base cases: if m == 0 then no solution for any sum
    for (int i = 0; i <= sum; i++) {
        csTable[i][0] = 0;
    }
    // base case: if sum = 0 then there is only one solution for any input set: just take none of each of the items.
    for (int j = 0; j <= m; j++) {
        csTable[0][j] = 1;
    }

    for (int i = 1; i <= sum; i++) {
        for (int j = 1; j <= m; j++) {
            // solutions excluding coins[j]
            final int s1 = csTable[i][j - 1];
            // solutions including coins[j]
            // look at the column index in csTable[i - coins[j - 1]][j]. This is not j-1 as we can use as much coin
            // of type j as we like.
            final int s2 = (i - coins[j - 1]) >= 0 ? csTable[i - coins[j - 1]][j] : 0;

            csTable[i][j] = s1 + s2;
        }
    }

    return csTable[sum][m];
}

Also refer to http://www.algorithmist.com/index.php/Coin_Change
Read full article from Dynamic Programming | Set 7 (Coin Change) | GeeksforGeeks

Dynamic Programming | Set 7 (Coin Change) | GeeksforGeeks

Recursive Formulation

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