(This article was first published on

**YGC » R**, and kindly contributed to R-bloggers)In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation: 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possible to make £2 in the following way: 1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p How many different ways can £2 be made using any number of coins?

Recursive algorithm is very straightforward.

^{?}View Code RSPLUS

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | coins <- c(1,2,5,10,20,50,100,200) findChange <- function(money, numCoins) { if (numCoins ==1) return(1) s <- 0 for (i in 1:numCoins) { remain <- money-coins[i] if(remain==0) s <- s+1 if(remain>0) s <- s+findChange(remain,i) } return(s) } print(findChange(200,length(coins))) |

> system.time(source("Problem31.R")) [1] 73682 user system elapsed 0.89 0.00 0.89

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