How do I handle rounding the purchase amount to the nearest multiple of 10?

You can use simple math to round the purchase amount to the nearest multiple of 10 by checking the difference with the nearest multiples of 10.

What should I do when the `purchaseAmount` is exactly 5?

When the amount is 5, round it to 10 because 5 is closer to 10 than to 0.

What happens if the `purchaseAmount` is 0 or 100?

If the amount is 0, the final balance remains 100. If the amount is 100, the final balance becomes 0.

What is the time complexity of the solution?

The time complexity is O(1) because you only need to round the number and subtract from the initial balance.

How does GhostInterview help with this problem?

GhostInterview helps by clearly outlining the solution steps and ensuring that you understand the rounding logic and edge cases in this simple mathematical problem.

#2806

Easy

auto_awesome数学·driven

LeetCode 题解工作台

取整购买后的账户余额

一开始，你的银行账户里有 100 块钱。给你一个整数 purchaseAmount ，它表示你在一次购买中愿意支出的金额。在一个商店里，你进行一次购买，实际支出的金额会向最近的 10 的倍数取整。换句话说，你实际会支付一个非负金额 roundedAmount ，满足 roundedA…

数学

题目描述

一开始，你的银行账户里有 100 块钱。

给你一个整数purchaseAmount ，它表示你在一次购买中愿意支出的金额。

在一个商店里，你进行一次购买，实际支出的金额会向最近的 10 的倍数取整。换句话说，你实际会支付一个非负金额 roundedAmount ，满足 roundedAmount 是 10 的倍数且 abs(roundedAmount - purchaseAmount) 的值最小。

如果存在多于一个最接近的 10 的倍数，较大的倍数 是你的实际支出金额。

请你返回一个整数，表示你在愿意支出金额为 purchaseAmount 块钱的前提下，购买之后剩下的余额。

注意： 0 也是 10 的倍数。

示例 1：

输入：purchaseAmount = 9
输出：90
解释：这个例子中，最接近 9 的 10 的倍数是 10 。所以你的账户余额为 100 - 10 = 90 。

示例 2：

输入：purchaseAmount = 15
输出：80
解释：这个例子中，有 2 个最接近 15 的 10 的倍数：10 和 20，较大的数 20 是你的实际开销。
所以你的账户余额为 100 - 20 = 80 。

提示：

0 <= purchaseAmount <= 100

lightbulb

解题思路

方法一：枚举 + 模拟

我们在 $[0, 100]$ 的范围内枚举所有的 $10$ 的倍数，然后找到与 purchaseAmount 最接近的那个数，记为 $x$ ，那么答案就是 $100 - x$ 。

时间复杂度 $O(1)$ ，空间复杂度 $O(1)$ 。

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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

speed

复杂度分析

指标	值
时间	complexity is O(1) because the operations consist only of rounding the number and performing a subtraction, both constant-time operations. The space complexity is O(1) as well since no additional data structures are required.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Can the candidate clearly explain the rounding concept?
question_mark
Does the candidate handle all edge cases correctly?
question_mark
Is the candidate able to recognize the simplicity of the solution and not overcomplicate the problem?

warning

常见陷阱

外企场景

error
Failing to round correctly, especially for values like 5 or 95 that are at the rounding threshold.
error
Overcomplicating the solution by using unnecessary loops or conditional structures.
error
Not correctly handling the boundary cases such as 0 or 100.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modify the problem to work with a different starting balance, such as 50 dollars, and adapt the solution accordingly.
arrow_right_alt
Implement a similar solution where the rounding is done to a multiple of 5 instead of 10.
arrow_right_alt
Extend the problem by adding a tax rate that must be deducted before rounding the purchase amount.

help

常见问题

外企场景

继续练习

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