How do you solve the Kids With the Greatest Number of Candies problem?

You solve it by checking if each kid's total candies, after adding extraCandies, is greater than or equal to the maximum candies among all the kids.

What pattern is used in the Kids With the Greatest Number of Candies problem?

The problem follows an array-driven solution pattern, where you manipulate and compare array values to solve the task.

Can multiple kids have the greatest number of candies in the Kids With the Greatest Number of Candies problem?

Yes, multiple kids can have the greatest number of candies if their total candies after receiving extraCandies match or exceed the maximum number.

What is the time complexity of the solution for Kids With the Greatest Number of Candies?

The time complexity is O(n), where n is the number of kids, because we scan the candies array twice: once to find the maximum and once to compare each kid's total.

What are the key steps in the solution to the Kids With the Greatest Number of Candies?

The key steps are finding the maximum candies, checking if each kid can reach that maximum after adding extraCandies, and returning the result as a boolean array.

#1431

Easy

auto_awesome数组·driven

LeetCode 题解工作台

拥有最多糖果的孩子

有 n 个有糖果的孩子。给你一个数组 candies ，其中 candies[i] 代表第 i 个孩子拥有的糖果数目，和一个整数 extraCandies 表示你所有的额外糖果的数量。返回一个长度为 n 的布尔数组 result ，如果把所有的 extraCandies 给第 i 个孩子之后，他会…

数组

题目描述

有 n 个有糖果的孩子。给你一个数组 candies，其中 candies[i] 代表第 i 个孩子拥有的糖果数目，和一个整数 extraCandies 表示你所有的额外糖果的数量。

返回一个长度为 n 的布尔数组 result，如果把所有的 extraCandies 给第 i 个孩子之后，他会拥有所有孩子中最多的糖果，那么 result[i] 为 true，否则为 false。

注意，允许有多个孩子同时拥有最多的糖果数目。

示例 1：

输入：candies = [2,3,5,1,3], extraCandies = 3
输出：[true,true,true,false,true] 
解释：如果你把额外的糖果全部给：
孩子 1，将有 2 + 3 = 5 个糖果，是孩子中最多的。
孩子 2，将有 3 + 3 = 6 个糖果，是孩子中最多的。
孩子 3，将有 5 + 3 = 8 个糖果，是孩子中最多的。
孩子 4，将有 1 + 3 = 4 个糖果，不是孩子中最多的。
孩子 5，将有 3 + 3 = 6 个糖果，是孩子中最多的。

示例 2：

输入：candies = [4,2,1,1,2], extraCandies = 1
输出：[true,false,false,false,false] 
解释：只有 1 个额外糖果，所以不管额外糖果给谁，只有孩子 1 可以成为拥有糖果最多的孩子。

示例 3：

输入：candies = [12,1,12], extraCandies = 10
输出：[true,false,true]

提示：

n == candies.length
2 <= n <= 100
1 <= candies[i] <= 100
1 <= extraCandies <= 50

lightbulb

解题思路

方法一

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class Solution:
    def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
        mx = max(candies)
        return [candy + extraCandies >= mx for candy in candies]

speed

复杂度分析

指标	值
时间	O(n)
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Can the candidate identify an array-driven solution strategy?
question_mark
Is the candidate able to recognize when an approach involves comparing elements in an array?
question_mark
Does the candidate optimize the solution to avoid unnecessary recalculations?

warning

常见陷阱

外企场景

error
Failing to find the maximum value before comparing each kid's total with extraCandies.
error
Not understanding the requirement that multiple kids can have the greatest number of candies.
error
Mistaking the problem as requiring sorting, when it can be solved with a simple comparison.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider varying the number of kids or the size of extraCandies.
arrow_right_alt
What if extraCandies were negative or zero?
arrow_right_alt
How would the solution change if we needed to find kids who would not have the greatest number of candies?

help

常见问题

外企场景

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