Group expense splitter for trips and parties

Expense splitter: enter who paid what to get each person's equal share and who owes whom, settled with the fewest possible payments.

Atualizado 28 de setembro de 2026Exemplos verificados: 4

Each person's share

Casas decimais: 2; Ao mais próximo; empates para o dígito par

Total spent: $195.50
Payments to settle up: 3
People: 4
Who pays whom: Jo pays Alex $48.88; Priya pays Alex $18.88; Sam pays Alex $3.38

4 people spent $195.50, so each owes $48.88. 3 payments settle everything: Jo pays Alex $48.88; Priya pays Alex $18.88; Sam pays Alex $3.38.

Paid minus share

Payments to settle up Linhas: 3

De	Para	Amount
Jo	Alex	$48.88
Priya	Alex	$18.88
Sam	Alex	$3.38

Como é calculado S

Total and equal share
$\frac{195.50}{4} = 48.875$
Balance: paid minus share
Alex 120 − 48.875 = 71.125; Sam 45.5 − 48.875 = −3.375; Priya 30 − 48.875 = −18.875; Jo 0 − 48.875 = −48.875
Fewest payments
4 people are owed or owe money. Grouping them into sets whose balances cancel exactly, each set of g people settles in g − 1 payments: 3 in total.
Amounts are rounded to the cent for display, so a payment can differ from the exact balance by less than a cent.

Sobre Group expense splitter for trips and parties

Each person's share is the total spent divided by the number of people. A balance is what someone paid minus that share: a positive balance is owed money and a negative one owes it. The calculator then splits the balances into as many groups as possible whose balances cancel out exactly, because a group of g people can always settle in g − 1 payments.

With the defaults, Alex, Sam and Priya paid 195.50 between them and Jo paid nothing, so each share is 48.875. Three payments settle everything: Jo pays Alex 48.88, Priya pays Alex 18.88 and Sam pays Alex 3.38.

Finding the fewest payments is NP-hard in general (Verhoeff, 2004), so the exact search runs when up to 16 people have a non-zero balance; larger groups fall back to paying the largest creditor from the largest debtor. Payments are shown to the cent, so each can differ from the exact balance by less than a cent.

Exemplos resolvidos

Three payers and one person who paid nothing

Who paid what: Alex paid 120 for groceries Sam paid 45.50 for fuel Priya paid 30 for snacks
Everyone sharing the costs: Alex, Sam, Priya, Jo

Total spent: 195.50
Each person's share: 48.88
Payments to settle up: 3
Who pays whom: Jo pays Alex $48.88; Priya pays Alex $18.88; Sam pays Alex $3.38

Fonte de verificação: 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

A case where matching the largest amounts first needs 4 payments, not 3

Who paid what: A paid 14 B paid 13 C paid 12 D paid 5 E paid 6

Each person's share: 10.00
Payments to settle up: 3

Fonte de verificação: Python brute-force partition into zero-sum groups {A, E}, {B, C, D}: 5 − 2 = 3; largest-first greedy gives 4

Everyone paid the same

Who paid what: A, 10 B, 10 C, 10

Each person's share: 10.00
Payments to settle up: 0
Who pays whom: Everyone is even — no payments needed.

Fonte de verificação: Definition: every balance is 0

A payer missing from the list is added

Who paid what: Alex paid 60 Sam paid 20 Kim paid 40
Everyone sharing the costs: Alex, Sam

People: 3
Each person's share: 40.00
Payments to settle up: 1
Who pays whom: Sam pays Alex $20.00

Fonte de verificação: Hand calculation: 120/3 = 40; Alex +20, Sam −20, Kim 0

Perguntas

How do you split expenses in a group?

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.

What is the fewest number of payments needed to settle up?

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.

Why can paying off the biggest debts first need extra payments?

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.

What if some costs are shared by only part of the group?

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.

Qual é a precisão de “Group expense splitter for trips and parties”?

A precisão depende dos dados inseridos e das hipóteses do método. O cálculo decimal usa 50 algarismos significativos, mas estimativas, métodos numéricos e dados de origem podem ter menor precisão; o arredondamento exibido não elimina essas limitações. Exemplos resolvidos verificados com fontes independentes: 4. Por exemplo, “Three payers and one person who paid nothing” é verificado com 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.

De onde vem o método?

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

Sobre esta calculadora

\text{share} = \frac{\sum \text{paid}}{n},\qquad \text{balance}_i = \text{paid}_i - \text{share},\qquad \text{payments} = k - \max(\text{zero-sum groups})

Fontes

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

Verificado com as referências

Esta calculadora inclui 4 exemplos resolvidos com respostas de fontes independentes. Eles fazem parte do conjunto de testes e você também pode executá-los aqui.