Is there a way to do this in O(n) time and O(1) auxiliary space ? Complexity of the solution isO(N*log(N)) For every indexiwe'll try and find the smallest length subarray such that it ends at positioniand it's sum is >= k.pref[i]is equal to the prefix sum of the array ...
So the idea is to replace each element of the given array by it's compressed value [index of this element if we sort the whole array]. It's simply done inO(n.log(n))n.log(n)). Also, save the inverse mapping, i.e.pos[compressedpos[compressedvalue(a[i])]=ivalue(a[i])]=i. ...
codeforces.com/contest/ 题意 给你一个长度为 n 的数组 a 。请找出一个长度至少为 k 且中位数最大的子数组 a[l..r]。 在长度为 n 的数组中,中位数是指按非递减顺序排序后,占据位置编号 ⌊n+12⌋ 的元素。例如 median([1,2,3,4])=2 , median([3,2,1])=2 , median([2,1,2,1])=...
class Solution { public: vector<int> GetLeastNumbers_Solution(vector<int> input, int k) { vector<int> res; if(k <= || k > input.size()) return res;int idx =; while(idx < k){ res.push_back(input[idx++]); } BubbleSort(res);...
Codeforces 1073G Yet Another LCP Problem SASA+单调栈 题意给出一个字符串ss和qq个询问。 每次询问给出两个长度分别为k,lk,l的序列aa和序列bb。 求∑ki=1∑lj=1lcp(s[ai…n],s[bj…n])∑i=1k∑j=1llcp(s[ai…n],s[bj…n])Solution...
Solution首先考虑简化版,如果边权都不一样怎么做。对于这样的经典模型,通用的做法是对边进行排序,按边权从小到大枚举边,这条边的得分为它连接的两个联通块的大小之积,计算之后用并查集合并两个连通块。现在的问题中存在权值相同的边权,显然我们每次枚举需要把权值相同的边统一处理了,难点在于计算得分策略的设计。
Codeforces-20152016-northwestern-european-regional-contest-nwerc-A题 一、题目 二、题意 (1)一开始理解成:它最多需要开多少台电脑。同时,我又有个疑问,既然是最多需要开多少台,那不变成了总共有几个人开几台是最大的结果。然后,WA了无数发。直到比赛结束。。。其实说到底还是不明白题目意思。 (2)真正的意思...
2019-12-04 22:35 − 1.全排序方法 class Solution: def kClosest(self, points, K): points.sort(key= lambda x: x[0]**2 + x[1]**2) return points[:K] 2. 堆排序... 流星小子 0 293 在上亿级别的数当中找前1000个数 2019-12-03 21:04 − 方法1: 对于这道题第一个想到的方...
It is faster than the previous solution.Solution Using Numeric OperationsTo get a k-digit number with all k in it, we can use this algorithm.First, create a k-digit number with all 1. Multiply that number with k. That gives us the number we want....
Thank you for your time. I think you are talking about using a merge sort tree, I tried using that but it exceeds time limit. Also, I am finding rank of k in array and to do that I am assigning it the rank of the last element that is not greater than k in the array, I think...