2303. Calculate Amount Paid in Taxes
Description
You are given a 0-indexed 2D integer array brackets
where brackets[i] = [upperi, percenti]
means that the ith
tax bracket has an upper bound of upperi
and is taxed at a rate of percenti
. The brackets are sorted by upper bound (i.e. upperi-1 < upperi
for 0 < i < brackets.length
).
Tax is calculated as follows:
- The first
upper0
dollars earned are taxed at a rate ofpercent0
. - The next
upper1 - upper0
dollars earned are taxed at a rate ofpercent1
. - The next
upper2 - upper1
dollars earned are taxed at a rate ofpercent2
. - And so on.
You are given an integer income
representing the amount of money you earned. Return the amount of money that you have to pay in taxes. Answers within 10-5
of the actual answer will be accepted.
Example 1:
Input: brackets = [[3,50],[7,10],[12,25]], income = 10 Output: 2.65000 Explanation: Based on your income, you have 3 dollars in the 1st tax bracket, 4 dollars in the 2nd tax bracket, and 3 dollars in the 3rd tax bracket. The tax rate for the three tax brackets is 50%, 10%, and 25%, respectively. In total, you pay $3 * 50% + $4 * 10% + $3 * 25% = $2.65 in taxes.
Example 2:
Input: brackets = [[1,0],[4,25],[5,50]], income = 2 Output: 0.25000 Explanation: Based on your income, you have 1 dollar in the 1st tax bracket and 1 dollar in the 2nd tax bracket. The tax rate for the two tax brackets is 0% and 25%, respectively. In total, you pay $1 * 0% + $1 * 25% = $0.25 in taxes.
Example 3:
Input: brackets = [[2,50]], income = 0 Output: 0.00000 Explanation: You have no income to tax, so you have to pay a total of $0 in taxes.
Constraints:
1 <= brackets.length <= 100
1 <= upperi <= 1000
0 <= percenti <= 100
0 <= income <= 1000
upperi
is sorted in ascending order.- All the values of
upperi
are unique. - The upper bound of the last tax bracket is greater than or equal to
income
.
Solution
calculate-amount-paid-in-taxes.py
class Solution:
def calculateTax(self, brackets: List[List[int]], income: int) -> float:
n = len(brackets)
first = min(brackets[0][0], income)
res = first * (brackets[0][1] / 100)
income -= first
for i in range(1, n):
k = brackets[i][0] - brackets[i - 1][0]
curr = min(income, k)
income -= curr
res += curr * (brackets[i][1] / 100)
if income == 0:
break
return res