hackerrank alternatives

Word, tool update-alternatives is the function of the management system under the similar function of the symbolic link, the way is to have a similar function of the program, not directly call the program, but with a unified symbolic link to call, as to the symbolic linkWhich of the similar function programs is pointed to, is configured and managed by the tool update-alternatives.It's very abstract, so let's take an example and see what's going on nat

Update-alternatives is a tool dedicated to maintaining system command links in Linux systems, which makes it easy to set which command to use by default, which software version, for example, we have both open jdk and Sun JDK Two versions installed in the system. And we want the system to use the sun JDK by default, what should we do? Through the update-alternatives can be very convenient to achieve.To view

Alternatives and update-alternatives seem to be the same tool (different for each Linux distribution), so here's an example of update-alternatives. Update-alternatives is a very useful tool in Linux, not only in the Debian series, but also in the Redhat series.Linux has grown to today, and there are so many available s

This is a very smart observation:http://www.martinkysel.com/hackerrank-sherlock-and-gcd-solution/#http://www.martinkysel.com/hackerrank-sherlock-and-gcd-solution/# fromFunctoolsImportReducedefgcd (A, b):ifA = = 1orb = = 1: return1ifA% b = =0:returnbreturnGCD (b, a%b)defMULTIPLE_GCD (arr):returnReduceLambdax,y:gcd (x, y), arr)#################T =int (input ()) forIinchRange (0,t): N=int (input ()) Arr

Yesterday for the first time to participate in a Hackerrank game, from 11 o'clock Noon to do the night 12 o'clock, finally too tired, at the end of the game one hour to go to bed. Completely passed the first two questions, the third question for a sentence has not been correctly understood, so did not pass the majority of test case, the final ranking in the contest in 23**/60**. Oneself is holding the mentality of study to go, although this achievemen

A Sly knapsack problem in disguise! Thanks to Https://github.com/bhajunsingh/programming-challanges/tree/master/hackerrank/algorithms/the-indian-jobLesson learnt: The Italian\indian job is two-way knapsack. And some complext problem can be converted to a simpler one.Code is amazingly simple:#include #include#include#include#includeusing namespacestd;intKnapsack (vectorint> a,intG) { intn =a.size (); Vectorint> dp (G +1); for(inti =1; I ) for(

Classic and challenging dp! And you need combine several tricks together with DP-make it 100% pass.My Main reference is here:https://github.com/havelessbemore/hackerrank/blob/master/algorithms/bit_manipulation/ String-transmission.javaBasically it is a counting problem.It's easy-to-possible count should be SUM (CR (n, 0), CR (n, 1),.. Cr (n, k)) without the limitation of "No Periodic strings". And then we can count how many "periodic" strings there is

update-alternatives--Management soft Connect/Word linked (Symbolic links) Command format update-alternatives [Options] Command Detailed description Update-alternatives can be created, deleted, repaired, soft-connected, and can display information about existing soft links, all of which constitute the Debian alternative system (

Alternatives is an important version management tool under UNIX, which was first seen in the Debain system, because its powerful usability makes it a large number of porting in various UNIX environments, And we probably most often use the Java OPENJDK Toolkit in the usual version of the conversion, and in the alternatives implementation of multi-version is the replacement system using the command link to im

, k, b);CIN >> N >> k >>b; //cout LL mi =0, mx =0; for(inti =1; I i) {mi+=i; } for(LL i = k; I >= K-b +1; --i) {mx+=i; //if (MX if(MX >= N) Break;//OK } //cout if(MX N) {printf ("-1\n"); return ; } if(Mi >N) {printf ("-1\n"); return ; } LL Now=mi; intto =b; LLGet=K; for(inti =1; I i; while(true) { now= Now-a[to] +Get; A[to]=Get; if(Now >=N) {LL cut= Now-N; A[to]-=cut; for(inti =1; I 1; ++i) {printf ("%lld", A[i]); } printf ("%lld\n", A[b]); Break; } to--; Get--; }}i

second by second. the problem is that the protein rings can be very long (up to 1 million proteins in a single ring) and they want to know the state of the ring after upto 109 seconds. thus their software takes too long to report the results. they ask you for your help. Input FormatInput contains 2 lines.First line has 2 integers n and k, n being the length of the protein ring and K the desired number of seconds.Second line contains a string of length N containing uppercase letters A, B, C or

Question Link Ashton appeared for a job interview and is asked the following question. arrange all the distinct substrings of a given string in lexicographical order and concatenate them. print the kth character of the concatenated string. it is assured that given value of K will be valid I. e. there will be a kth character. can you help Ashton out with this? Input FormatFirst line will contain a number t I. e. Number of test cases.First line of each test case will contain a string containing

Similar Pair _ HackerRank, pairhackerrank Hacker rank is really more difficult than leetcode .. This question is a bit clever .. Each path is searched in depth and then enumerated. Most people think about it, but the key is that this definitely times out. The trick is to create a line segment tree for each path to accelerate the query. The complexity of each similar query changes from O (h) to O (lgh ).. Two mistakes were made. (1) Use long to store

Greedy beats DP this time... I tried several DP solutions first, but all failed with re \ TLE. If you 'feel' the problem, greedy shocould be working: (A solution from discussion) Def getiterator (n): While n> 0: If n % 3 = 0: break; else: N-= 5 return n t = input () for I in xrange (t): n = int (input () rows = getrows (n) If rows Hackerrank-Sherlock and the beast

Another fun greedy problem to work on:we simply go from first to second last person, as long someone is odd, we distribut E bread to she and her next.#include #includeusing namespacestd;intMain () {intN; CIN>>N; Vectorint>B (N); for(intB_i =0; B_i ) {cin>>B[b_i]; } intCNT =0; for(inti =0; I 1; i + +) { if(B[i]%2) {B[i]++; B[i+1]++; CNT+=2; } } if(B.back ()%2) cout"NO"Endl; ElsecoutEndl; return 0;}Hackerrank "Fair rations"

An intuitive Prim algorithm impl.#include #include#include#include#includeusing namespacestd;structedge{Edge (): s (0), T (0), D (0{} Edge (unsigned RS, unsigned rt, unsigned rd): s (rs), T (RT), D (RD) {} unsigned s; unsigned t; unsigned D; BOOL operator()(ConstEdge e1,ConstEdge E2) { returne1.d >e2.d; }};intMain () {LongN, M; CIN >> N >>m; //From -to-and (length, id)unordered_mapG; for(inti =0; I ) {Unsigned A, B, D; CIN>> a >> b >>D; G[A][B]= G[b][a] =D; } unsigned s; CIN>>s; Un

"How many inverted pairs"-so usually ends up with mergesort solution (of course there is other solutions out there)defmergesort (arr):ifLen (arr) = = 1: return0, arr mid= Len (arr)//2Cnt1, arr1=mergesort (Arr[:mid]) Cnt2, arr2=mergesort (arr[mid:])#RET =[] CNT=0 I=0 J=0 Inx=0 whileI andJ Len (arr2):ifArr1[i] Arr2[j]: Ret.append (arr1[i]) I+ = 1elifARR2[J] Arr1[i]: Ret.append (arr2[j]) CNT+ = Len (arr1)-I#arr2-arr1J + = 1 whileI Len (arr1): Ret.append (Arr1[i]) I+ = 1 whileJ Len (arr2): R

I caught the sparkle in my mind and got AC1! It is a great great experience!So the basic Idea:permute on 3 consecutive items doesn ' t change the parity of the No. of inversions. Each permutation can is only remove 0 or 2 inversions. So we say "YES" when No. of inversion% 2 = = 0. And we use MergeSort to count it in O (NLGN).RET =0defmerge (arr1, arr2):Globalretif notARR1:returnarr2if notARR2:returnarr1 forV2inchArr2:arr1.append (v2) I= Len (arr1)-1 while(I >0):ifArr1[i] ]: Arr1[i-1],arr1[i] =

Longest Common subsequence in disguise. Python Impl. Have TLE in one case. and C + + is fine.#include #include#include#include#includestring>#include#includeusing namespacestd;intMain () {stringA, B; CIN>>A; CIN>>b; size_t Len=a.length (); Vectorint>> DP (len +1, vectorint> (len +1,0)); for(inti =0; I ) for(intj =0; J ) { if(I * j = =0) Dp[i][j] =0; Else if(A[i-1] = = B[j-1]) Dp[i][j]= Dp[i-1][j-1] +1; ElseDp[i][j]= Max (Dp[i-1][J], Dp[i][j-1]); } coutDp[len][len]; return 0;}

This is the classic LCS problem. Since it requires you-to-print one longest common subsequence, just use the O (m*n)-space version here.My accepted code is as follows.1#include 2#include 3#include 4 5 using namespacestd;6 7vectorint> LCS (vectorint> A, vectorint>b) {8 intn = a.size (), M =b.size ();9vectorint> > DP (n +1, vectorint> (M +1,0));Ten for(inti =1; I ) { One for(intj =1; J ) { A if(A[i-1] = = B[j-1]) Dp[i][j] = dp[i-1][j-1] +1; - ElseDP[I][J]