Find the least number of coins required that can make any change from 1 to 99 cents

Question

Recently I challenged my co-worker to write an algorithm to solve this problem:

Find the least number of coins required that can make any change from

Answer 1

Here goes my solution, again in Python and using dynamic programming. First I generate the minimum sequence of coins required for making change for each amount in the range 1..99, and from that result I find the maximum number of coins needed from each denomination:

def min_any_change():
    V, C = [1, 5, 10, 25], 99
    mxP, mxN, mxD, mxQ = 0, 0, 0, 0
    solution = min_change_table(V, C)
    for i in xrange(1, C+1):
        cP, cN, cD, cQ = 0, 0, 0, 0
        while i:
            coin = V[solution[i]]
            if coin == 1:
                cP += 1
            elif coin == 5:
                cN += 1
            elif coin == 10:
                cD += 1
            else:
                cQ += 1
            i -= coin
        if cP > mxP:
            mxP = cP
        if cN > mxN:
            mxN = cN
        if cD > mxD:
            mxD = cD
        if cQ > mxQ:
            mxQ = cQ
    return {'pennies':mxP, 'nickels':mxN, 'dimes':mxD, 'quarters':mxQ}

def min_change_table(V, C):
    m, n, minIdx = C+1, len(V), 0
    table, solution = [0] * m, [0] * m
    for i in xrange(1, m):
        minNum = float('inf')
        for j in xrange(n):
            if V[j] <= i and 1 + table[i - V[j]] < minNum:
                minNum = 1 + table[i - V[j]]
                minIdx = j
        table[i] = minNum
        solution[i] = minIdx
    return solution

Executing min_any_change() yields the answer we were looking for: {'pennies': 4, 'nickels': 1, 'dimes': 2, 'quarters': 3}. As a test, we can try removing a coin of any denomination and checking if it is still possible to make change for any amount in the range 1..99:

from itertools import combinations

def test(lst):
    sums = all_sums(lst)
    return all(i in sums for i in xrange(1, 100))

def all_sums(lst):
    combs = []
    for i in xrange(len(lst)+1):
        combs += combinations(lst, i)
    return set(sum(s) for s in combs)

If we test the result obtained above, we get a True:

test([1, 1, 1, 1, 5, 10, 10, 25, 25, 25])

But if we remove a single coin, no matter what denomination, we'll get a False:

test([1, 1, 1, 5, 10, 10, 25, 25, 25])