2806. Account Balance After Rounded Purchase 

Description

Initially, you have a bank account balance of 100 dollars.

You are given an integer purchaseAmount representing the amount you will spend on a purchase in dollars, in other words, its price.

When making the purchase, first the purchaseAmount is rounded to the nearest multiple of 10. Let us call this value roundedAmount. Then, roundedAmount dollars are removed from your bank account.

Return an integer denoting your final bank account balance after this purchase.

Notes:

0 is considered to be a multiple of 10 in this problem.
When rounding, 5 is rounded upward (5 is rounded to 10, 15 is rounded to 20, 25 to 30, and so on).

Example 1:

Input: purchaseAmount = 9

Output: 90

Explanation:

The nearest multiple of 10 to 9 is 10. So your account balance becomes 100 - 10 = 90.

Example 2:

Input: purchaseAmount = 15

Output: 80

Explanation:

The nearest multiple of 10 to 15 is 20. So your account balance becomes 100 - 20 = 80.

Example 3:

Input: purchaseAmount = 10

Output: 90

Explanation:

10 is a multiple of 10 itself. So your account balance becomes 100 - 10 = 90.

Constraints:

0 <= purchaseAmount <= 100

Solutions

Solution 1: Enumeration + Simulation

We enumerate all multiples of 10 within the range $[0, 100]$, and find the one that is closest to purchaseAmount, denoted as $x$. The answer is $100 - x$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def accountBalanceAfterPurchase(self, purchaseAmount: int) -> int:
        diff, x = 100, 0
        for y in range(100, -1, -10):
            if (t := abs(y - purchaseAmount)) < diff:
                diff = t
                x = y
        return 100 - x

Java

class Solution {
    public int accountBalanceAfterPurchase(int purchaseAmount) {
        int diff = 100, x = 0;
        for (int y = 100; y >= 0; y -= 10) {
            int t = Math.abs(y - purchaseAmount);
            if (t < diff) {
                diff = t;
                x = y;
            }
        }
        return 100 - x;
    }
}

C++

class Solution {
public:
    int accountBalanceAfterPurchase(int purchaseAmount) {
        int diff = 100, x = 0;
        for (int y = 100; y >= 0; y -= 10) {
            int t = abs(y - purchaseAmount);
            if (t < diff) {
                diff = t;
                x = y;
            }
        }
        return 100 - x;
    }
};

Go

func accountBalanceAfterPurchase(purchaseAmount int) int {
	diff, x := 100, 0
	for y := 100; y >= 0; y -= 10 {
		t := abs(y - purchaseAmount)
		if t < diff {
			diff = t
			x = y
		}
	}
	return 100 - x
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

function accountBalanceAfterPurchase(purchaseAmount: number): number {
    let [diff, x] = [100, 0];
    for (let y = 100; y >= 0; y -= 10) {
        const t = Math.abs(y - purchaseAmount);
        if (t < diff) {
            diff = t;
            x = y;
        }
    }
    return 100 - x;
}