Can you solve this real interview question? Maximum Number of Coins You Can Get - There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows: * In each step, you will choose any 3 piles of coins (not necessari
In each step, you will choose any 3 piles of coins (not necessarily consecutive). Of your choice, Alice will pick the pile with the maximum number of coins. You will pick the next pile with maximum number of coins. Your friend Bob will pick the last pile. Repeat until there are no m...
You will pick the next pile with maximum number of coins. Your friend Bob will pick the last pile. Repeat until there are no more piles of coins. Given an array of integerspileswherepiles[i]is the number of coins in theithpile. Return the maximum number of coins which you can have. ...
输入:coins = [1,3]输出:2解释:你可以得到以下这些值: - 0:什么都不取 [] - 1:取 [1] 从 0 开始,你可以构造出 2 个连续整数。 示例2: 输入:coins = [1,1,1,4]输出:8解释:你可以得到以下这些值: - 0:什么都不取 [] - 1:取 [1] - 2:取 [1,1] - 3:取 [1,1,1] - 4:取 ...
447. Number of Boomerangs # 题目# Given n points in the plane that are all pairwise distinct, a “boomerang” is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters). Find the number of boomeran...
题目地址:https://leetcode.com/problems/number-of-segments-in-a-string/#/description 题目描述 Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printabl...
first coin will sum up to 2, and thus make a sum of 2 with the help of only 2 coins. This is the best and only solution for sum 2. Now we proceed to sum 3. We now have 2 coins which are to be analyzed - first and second one, having values of 1 and 3. Let's see the ...
1. Description Maximum Number of Coins You Can Get 2. Solution 解析:Version 1,先对数组排序,然后使用贪心算法,每次选两个最大的,选一个最小的,这样保证可以拿到第二大的,同时将最小值从候选数据中剔除。 Version 1 classSolution:defmaxCoins(self,piles:List[int])->int:piles.sort()maximum=0i=0j...
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