vector<vector<int>> dp(n, vector<int>(n, -1));intmyBest = utill(nums, dp,0, n-1);return2*myBest >= accumulate(nums.begin(), nums.end(),0); }intutill(vector<int>& v, vector<vector<int>> &dp,inti,intj){if(i > j)return0;if(dp[i][j] != -1)returndp[i][j];int...
128th LeetCode Weekly Contest Pairs of Songs With Total Durations Divisible by 60 In a list of songs, thei-th song has a duration oftime[i]seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by60. Formally, we want the number of indicesi...
关联问题 换一批 LeetCode Biweekly Contest 36 中有哪些有趣的题目? 在LeetCode Biweekly Contest 36 中,有哪些题目涉及到数据结构的应用? LeetCode Biweekly Contest 36 的时间限制是多少? 0. 赛后总结 昨天本来身体有点不舒服,就借口颓废了一整天,除了看剧啥事没干,想着比赛就算了吧,结果晚上还是没能逃过...
leetcode weekly contest 414 近期做的比较好的比赛了,还算是比较有难度的题目,还不错基本都 Ak 了。 3280. 将日期转换为二进制表示 给你一个字符串 date,它的格式为 yyyy-mm-dd,表示一个公历日期。 date 可以重写为二进制表示,只需要将年、月、日分别转换为对应的二进制表示(不带前导零)并遵循 year-mont...
LeetCode笔记:Weekly Contest 301 1. 题目一 给出题目一的试题链接如下: 2335. Minimum Amount of Time to Fill Cups 1. 解题思路 这一题其实仔细分析一下的话答案是比较明显的,就是总数的一半。 但是其中存在一个特例,就是如果某一个元素的个数大于总数的一半时,那么它就是答案。
每周参加weekly contest，发现前几名都会在半小时左右完成4题，我感觉对于一道陌生的MEDIUM题目，理解+思考...
Explanation: All three pairs have a total duration of 120, which is divisible by 60. 1. 2. 3. Note: 1 <= time.length <= 60000 1 <= time[i] <= 500 本来是个很简单的题目,不小心我给写复杂了。 但这样就很好理解了。我们找的就是余数也是i和60-i这种,然后考虑排列组合就好了。注意0和30...
Welcome toBiweekly Contest 43! Feel free to share and post your contest experience here! You can also view the rankings for the contesthere. Links to the individual problems are included below: Calculate Money in Leetcode Bank(3 points) ...
Algorithms Exercise: LeetCode Problems, LeetCode Weekly Contest etc. github.com/caipengbo/LeetCode-CPP Resources Readme Activity Stars 56 stars Watchers 6 watching Forks 17 forks Report repository Releases No releases published Packages No packages published Languages Java 99.9% C++ 0.1...
113th LeetCode Weekly Contest Largest Time for Given Digits,Givenanarrayof4digits,returnthelargest24hourtimethatcanbemade.Thesmallest24hourtimeis00:00,andthelargestis23:59.Startingf