Greedy Coin Change Time Complexity

Question

I'm trying to figure out the time complexity of a greedy coin changing algorithm. (I understand Dynamic Programming approach is better for this problem but I did that already). I'm not sure how to go about doing the while loop, but I do get the for loop.

I have the following where D[1...m] is how many denominations there are (which always includes a 1), and where n is how much you need to make change for.

This is my algorithm:

CoinChangeGreedy(D[1...m], n)
    numCoins = 0
    for i = m to 1
        while n ≥ D[i]
            n -= D[i]
            numCoins += 1
    return numCoins

`I understand DP is better` What is DP in this context? – Nov 13 '18 at 21:30 — , Nov 13 '18 at 21:30
@Amy Dynamic programming I'd guess? – CollinD Nov 13 '18 at 21:33 — CollinD, Nov 13 '18 at 21:33
Yes, DP was dynamic programming. Sorry for the confusion. – RiseWithMoon Nov 14 '18 at 04:30 — RiseWithMoon, Nov 14 '18 at 04:30

score 4 · Accepted Answer · answered Nov 13 '18 at 21:37

4

Let look at the edge cases.

At the worse case D include only 1 element (when m=1) then you will loop n times in the while loop -> the complexity is O(n).

If m>>n (m is a lot bigger then n, so D has a lot of element whom bigger then n) then you will loop on all m element till you get samller one then n (most work will be on the for-loop part) -> then it O(m).

Button line: O(max(m,n))

Hope that helps!

answered Nov 13 '18 at 21:37

dWinder

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I guess I don't understand how you loop `n` times when n is decremented by a value that is dependent in the denominations `D[1...m]` array. – RiseWithMoon Nov 14 '18 at 04:45
1

For the complexity I looked at the worse case - if `m=1` then it will loop `n` times (try to run example). Of course if `m` bigger then the complexity decrease. – dWinder Nov 14 '18 at 04:55
Okay that makes sense. What would the best-case be then? – RiseWithMoon Nov 14 '18 at 05:16
@dgfrey102790 O(1) when `m=n`. I guess average case will be little bit more then `n/m` when `m – dWinder Nov 14 '18 at 05:58
Would it be wrong to say `O(m + n)` instead of `O(max(m, n))`? – qz- Feb 03 '22 at 00:38
If we limit denominations to integers then the smallest possible denomination is `m`, and I think we can get a tighter bound from that and say `O(m + n / m)` – qz- Feb 03 '22 at 00:42

RiseWithMoon · Answer 2 · 2018-11-14T22:34:19.440

1

Thanks for the help. I changed around the algorithm I had to something I could easily calculate the time complexity for. Here's what I changed it to:

CoinChangeGreedy(D[1...m], n)
    numCoins = 0
    for i = m to 1
        if n/D[i] ≥ 1
            numCoins = numCoins + (n/D[i])
            n = n - [(n/D[i]) * D[i]]
    return numCoins

Where I calculated this to have worst-case = best-case \in \Theta(m)

edited Nov 14 '18 at 22:34

answered Nov 14 '18 at 22:23

RiseWithMoon

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Greedy Coin Change Time Complexity

2 Answers2