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Problem: given a set of n coins of unique face values, and a value change, find number of ways of making change for change.

Assuming we can use a denomination more than once, here's the pseudocode I came up with

1. NUM-WAYS(denom[], n, change)
2.   dp = [change + 1][n + 1]
3.   for i = 0 to n
4      dp[i][0] = 1
5.   xs = denom.sorted

6.   for i = 1 to change
7.     for j = 1 to n
8.       if xs[j - 1] > i
9.         dp[i][j] = dp[i][j - 1]
10.      else
11.        dp[i][j] = dp[i - xs[j - 1]][j] + dp[i][j - 1]

12.  return dp[change][n]

The above algorithm is clear to me. However, if we are only allowed to use a denomination once, then line 11 changes to dp[i - xs[j - 1]][j - 1] + dp[i][j - 1], as if we are not allowed to use the current denomination at all. I'm failing to wrap my head around this. Can you explain this?

Here're some test runs:

Change: 3, denominations: [8, 3, 1, 2]
11111
01111
01222
01233

// use once
Change: 3, denominations: [8, 3, 1, 2]
11111
01111
00111
00122

Change: 4, denominations: [3, 1, 2]
1111
0111
0122
0123
0134

// use once
Change: 4, denominations: [3, 1, 2]
1111
0111
0011
0012
0001
Abhijit Sarkar
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1 Answers1

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Let dp[i][j] denote the solution to the [i, j]-th subproblem; that is, dp[i][j] is the number of ways to make change for the amount i, using coins j through n.

Coin Change Problem with Repetition

  • The first case assumes that a single coin of the j-th denomination was taken. Since there's no constraint on the denominations, j may remain the same so it can be used again for smaller subproblems: dp[i - xs[j - 1]][j].

Coin Change Problem without Repetition

It's the same as the above problem with an additional constraint that a coin of some denomination can be only taken once.

  • The first case assumes that a single coin of the j-th denomination was taken. Since we can't use denomination j anymore, j changes to j-1: dp[i - xs[j - 1]][j - 1]

The second case is the same in both problems where the j-th denomination is ignored.

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Osama Fathy
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