hackerrank vs leetcode

Design and implement a data structure for Least recently Used (LRU) cache. It should support the following operations: get and set .get(key)-Get The value ('ll always be positive) of the key if the key exists in the cache, otherwise return-1.set(key, value)-Set or insert the value if the key is not already present. When the cache is reached its capacity, it should invalidate the least recently used item before inserting a new item.Basic ideas:Work with a map to find the problem.Use list maintena

termination condition?-When a or B is empty, return directly to A[k-1] or b[k-1]-When k = 1 o'clock, return min (a[0], b[0])//1th small represents the first element-When A[K/2-1] = = B[K/2-1], return to A[K/2-1] or B[K/2-1]C Language ImplementationStaticintFind_kth (intAintMintAintNintk); Staticint min(p, Q) {return(P DoubleFindmediansortedarrays (int* Nums1,intNums1size,int* NUMS2,intNums2size) {intm = nums1size;intn = nums2size;intTotal = M+n;intK = total/2;if(Total 0x01) {returnFind_kth (Num

Title: https://leetcode.com/problems/3sum-closest/"Label" Array; Pointers"Personal Analysis"This question and its sister problem 3Sum very similar, will not say more, specific analysis can refer to [leetcode][015] 3Sum1 Public classSolution {2 Public intThreesumclosest (int[] Nums,inttarget) {3 intresult =Target;4 intLen =nums.length;5 if(Len ) {6 return0;7 }8 Arrays.sort (nums);9 for(inti = 0

GivenNnon-negative integersa1 ,a2 , ...,an , where each represents a point at coordinate (I,ai ).NVertical lines is drawn such that the both endpoints of lineIis at (I,ai ) and (I, 0). Find lines, which together with X-axis forms a container, such that the container contains the most water.Note:you may not slant the container.This problem is very violent, equivalent to find two "planks", of which the short height of that * two between the distance. So our first thought was the Brute force O (n^2

Description:Count the number of prime numbers less than a non-negative number, nClick to show more hints.Credits:Special thanks to @mithmatt for adding this problem and creating all test cases.For all primes within N, previously seen a topic, by marking out all non-primes, and then finding the prime number, the code is as follows:1 Public intCountPrimes (intN) {2 if(n = = 0 | | n = = 1 | | n = = 2)3 return0;4 int[] flag =New int[n];5 for(inti = 2; I ) {6

. The distance difference between the elements on each row adjacent to the vertical column is a period, for example, when n=4, the period is 2*4-2 = 6, the element 0 and 6 are 6, the elements 1 and 7 differ by 6.In each row in the middle, there is an oblique element in the middle of each of the two vertical columns, for example, 1 and 7 are 5,7 and 13 in the middle is 11. The middle oblique element differs from the element on the previous vertical column to a period of -2* row index. As can be u

Leetcode Address: https://leetcode.com/problems/maximum-gap/description/Title DescriptionGiven an unordered array, find the maximum interval between the sorted elements.It is required to be implemented with linear time complexity.For example: The input is "3,6,9,1,10" and the output should be the maximum interval of 3.Problem analysisLinear time complexity implementation, it means that the element cannot be sorted first, because the complexity of this

Python implementation of Leetcode Primary algorithm--string# invert string def reversestring (s): Return s[::-1]# reverse number def reverse (x): if x Python implementation of the Leetcode primary algorithm--string

of the Top-down, there is redundancy in the space, each time need to repeat the results (such as the first behavior convenient second row calculation, need to save two values), and the need for special processing of the tail-end elements. So consider using bottom-up dynamic programming, starting from the last layer to calculate, set Dp[i][j] for the first layer J elements as the starting point of the smallest path and dynamic programming equation is:DP[I][J] = value[i][j] + min (dp[i-1][j], dp[

1 PrefaceThe reason for the beginning of the brush Leetcode algorithm problem, one is to face the autumn strokes, the 2nd is because the ability to improve their programming, in the blog to brush 5 or so on the algorithm of the problem, adhere to two months, hope Bo friends to supervise.This series intends to use C # and Java programming, why with two languages, because of the spring strokes, found that the company is in the C # is too few, only learn

Given an unsorted array return whether a increasing subsequence of length 3 exists or not in the array.Formally the function should: Return True if there exists I, J, K such that Arr[i] arr[j] arr[k] given 0≤ i J k ≤ n-1 else return false. Your algorithm should run in O (n) time complexity and O (1) space complexity.Examples:Given [1, 2, 3, 4, 5] ,Return true .Given [5, 4, 3, 2, 1] ,Return false .This problem and Leetcode