Group expense splitter for trips and parties

Add up everything spent, divide by the number of people to get the equal share, then subtract that share from what each person paid. For 195.50 spent across four people the share is 48.875: Alex, who paid 120, is owed 71.125, and Jo, who paid nothing, owes 48.875. Everyone with a negative balance pays someone with a positive one.

At most one fewer than the number of people with a non-zero balance, and fewer when the balances split into groups that cancel out, since each group of g people needs g − 1 payments. Tom Verhoeff's 2004 paper shows that finding the minimum is at least as hard as the subset-sum problem, so an exact answer needs a search.

It can miss groups that cancel out. With balances of +4, +3, +2, −5 and −4 (people A to E in the worked example), matching the largest amounts first takes 4 payments. Splitting them into {+4, −4} and {+3, +2, −5} settles the first pair in 1 payment and the trio in 2, so 3 in total.

This splitter shares every cost equally among everyone listed. For a cost that only some people share, such as two people's train tickets, run it separately with just those people, then add the payments from both runs. The combined list settles every debt, though it may use one or two more payments than a single optimised plan.

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 4 worked examples whose answers come from independent sources; for example, “Three payers and one person who paid nothing” is checked against Python decimal: 195.50/4 = 48.875; balances −48.875, −18.875, −3.375 rounded half-even to cents; 3 payments by brute-force subset search.

Verhoeff, T. (2004). Settling multiple debts efficiently: an invitation to computing science. Informatics in Education 3(1).