You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.
You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
Given scores of N athletes, find their relative ranks and the people with the top three highest scores, who will be awarded medals: “Gold Medal”, “Silver Medal” and “Bronze Medal”.
Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.
Examples:
[2,3,4] , the median is 3
[2,3], the median is (2 + 3) / 2 = 2.5
Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.
For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
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Window position Median
--------------------
[13-1]-353671
1[3-1-3]5367-1
13[-1-35]367-1
13-1[-353]673
13-1-3[536]75
13-1-35[367]6
Therefore, return the median sliding window as [1,-1,-1,3,5,6].
Note:
You may assume k is always valid, ie: k is always smaller than input array’s size for non-empty array.
Solution 0: Brute Force
Time complexity: O(n*klogk) TLE 32/42 test cases passed
Solution 1: Insertion Sort
Time complexity: O(k*logk + (n – k + 1)*k)
Space complexity: O(k)
C++ / vector
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// Author: Huahua
// Running time: 99 ms
classSolution{
public:
vector<double>medianSlidingWindow(vector<int>& nums, int k) {