package _interview_question /** * https://leetcode.com/discuss/interview-question/275785/facebook-phone-screen-count-subsets * * Input: Given an array A of -positive -sorted -no duplicate -integer A positive integer k Output: Count of all such subsets of A, Such that for any such ...
但是,它可以在O(nlogn)的复杂度中完成。 设S(n)表示前n个元素的和,SubSet(i,j)表示数组i到j的子集的和;给定的范围是[low,up]; 则简单来讲,要找满足low<=SubSet(i,j)<=up的SubSet的个数。 SubSet(i,j) = S(j) - S(i);也就是要求low<=S(j) - S(i)<=up中ij的个数。 这是可以看出必...
知道这个规律以后,可以累积上面的结果,把结果直接存在数组里面,暴力打表即可。O(1) 的时间复杂度。 参考代码 packageleetcode// 暴力打表法funccountNumbersWithUniqueDigits1(nint)int{res:=[]int{1,10,91,739,5275,32491,168571,712891,2345851,5611771,8877691}ifn>=10{returnres[10]}returnres[n]}// 打表...
length]); TreeSet<Integer> tset = new TreeSet<>(); int n = nums.length; for(int i = n - 1; i >= 0; i--) { TreeSet<Integer> subset = (TreeSet<Integer>)tset.headSet(nums[i]); counts.set(i, subset.size()); tset.add(nums[i]); } return counts; } } 正确 class Solu...
【leetcode】327. Count of Range Sum 题目如下: 解题思路:本题是560. Subarray Sum Equals K的升级版,可以参见560的解题思路。唯一的区别是560只给了一个精确的和K,而本题是给了一个和的范围,所以最终计数的时候遍历一下题目要求的区间即可。 代码如下:...
Input: nums = [1] Output: 1 Explanation: There is 1 square-free subset in this example: - The subset consisting of the 0th element [1]. The product of its elements is 1, which is a square-free integer. It can be proven that there is no more than 1 square-free subset in the ...
A set of practice note, solution, complexity analysis and test bench to leetcode problem set - leetcode/CountSubArrayFixBound.drawio at b58bcceb0ea27d0756ad72fb6a64b3b547fae221 · brianchiang-tw/leetcode