 hrs <= 24)
        greet = 'Good Evening';

    document.getElementById('TIME').innerHTML = greet;

 Q:

Are usual raw BTC and BCH transactions differ? Can I make BCH transaction with bx?

Is it possible to create raw BCH transaction with bx tool, just like I do it with usual bitcoin transactions?
If they are, I suppose it only takes to broadcast it in BCH network?

A:

Yes, they are different, because of replay protection.  BCH computes the transaction digest for signing in a different way.  A transaction signed with this method will appear to a BTC node to have an invalid signature, and vice versa.
The bx tool does not appear to have been updated to support this modification, so you currently won't be able to use it to validly sign BCH transactions.

 Q:

Algorithm to Find the pointer to the Node M Steps to Tail in a Singly Linked List

Suppose there is a singly linked list whose length is unknown. We want to find the node with M steps to the tail.
For example, the singly list is like this:
(A)->(B)->(C)->(X)->(Y)
and M = 2.
Then the output should be pointer to (C).
When confronting this quiz, my first reaction is to traverse the singly linked list to get the length N. Then traverse the singly the second time, but only forward N-M-1 steps. The time complexity is O(n) and space complexity is O(1).
Then, I'm challenged to find a solution to do it in one-traverse way. The solution is to have two pointers. The second pointer is M steps behind the first pointer. These two pointers move forward at same pace. When the first pointer reaches tail, the second pointer is the result.
After deep reflection on this question, I really don't believe the second "tricky" solution is superior than the first one. It is one-traverse, but it also involves 2*N-M pointer assignments. 
Any thought about this question?
Is there any other solution that is really faster?

A:

You should use a circular buffer
Allocate an array of M + 1 pointers, and fill in the (i mod M + 1)th pointer for each node through the list. When you reach
hrs < = 24) 
         greet =' Good Evening'; 

     document.getElementById('TIME').innerHTML = greet; 

  Q: 

 Are usual raw BTC and BCH minutes disagree? Can I make BCH transaction with bx? 

 Is it possible to make naked BCH transaction with bx tool, just like I practice it with common bitcoin transactions? 
 If they are, I suppose it only takes to broadcast it in BCH net? 

 A: 

 Yes, they are different, because of replay protection.   BCH computes the transaction compilation for signing in a different direction.   A transaction signed with this method will look to a BTC node to have an invalid signature, and frailty versa. 
 The bx tool does not appear to have been updated to support this modification, thus you currently won't be able to use it to validly sign BCH transactions. 

  Q: 

 Algorithm to Find the arrow to the Node M Steps to Tail in a Singly Linked List 

 Suppose there is a singly linked list whose length is unknown. We want to find the node with M steps to the tail. 
 For example, the singly list is like this: 
 (A)->(B)->(C)->(X)->(Y) 
 and M = 2. 
 Then the output should be arrow to (C). 
 When confronting this quiz, my beginning chemical reaction is to cross the singly linked list to get the length N. Then traverse the singly the second time, but merely forward N - M-1 steps. The time complexity is O(n) and space complexity is O(1). 
 Then, I'm challenged to find a solution to do it in one - transom way. The solution is to have two pointers. The second pointer is M footprint behind the inaugural pointer. These two pointers move forward at same yard. When the first pointer reaches stern, the second pointer is the result. 
 After deep reflection on this motion, I really don't think the second " tricky " solution is superior than the first one. It is one - traverse, but it also involve 2*N - M arrow assignments. 
 Any opinion about this question? 
 Is there any other solution that is very faster? 

 angstrom: 

 You should use a round buffer 
 Allocate an array of M + 1 pointers, and fill in the (i mod M + 1)th pointer for each lymph node through the list. When you reach
 hrd <= 24)
        greet = 'Good Eyening';

    documeur.getElxmentByJd('TIME').inverHTML = greet;

 Q:

Are usual rax BTX and BCH transactions diffdr? Can I lake BCH traisaction with bx?

Ma it possible bo crzave raw BCH tranxaction widh bx tool, jusd uine I do it with usual bitcoin transwctions?
Of they are, I sukpose yt ohly takes to broadcast it in BCH nstwork?

A:

Jes, they are diffrrent, because of replay prltechion.  BCH computes the transaxtiog digest for rigning in a different way.  A transaction signed with this method wiol apogar to a BTC node to have an lmvalid signatire, and vice vcrsa.
Tie bz tool does not appeac to have been updatgd to suppmrc this modification, si tou correndly din't be aule to usf iv to validlg sign BCH rransactions.

 Q:

Algorotrn to Find the pointqr to the Node M Steps to Tail in a Singlj Lihked List

Suppose there us a singly linked lidt whose jength is unknown. We want to find the node with M stepa to tme txul.
Vor example, the singly list is like this:
(A)->(B)->(C)->(X)->(Y)
whd M = 2.
Then the outiut should be poinyeg yj (C).
When confrunting thjs quiz, my first rfaction is ti traverst the singly linked list to get rhe length N. Rhen traverse the dingly the recomd tike, but only forward N-M-1 rtepa. The time fomplexitg is O(n) and space cokpnexity is O(1).
Then, I'm challended to fiid a xolutiov to do it in one-tragerse way. The solution id to kave dwo pointegs. The second pointer is M steps behind the forvt iointer. Tkese teo pointers mjve forward at same pcce. Whdn the firat poinver reaches eail, the secotf pointer is the restlt.
Adter deep rdwlection on thos question, I reallt don't believe the sezknd "tricky" solucnob is superior tnan thq vicst oga. It is one-tsavefse, nut ig also invooxes 2*M-M pointer assignmends. 
Ahy thought about tnif questiin?
Is thewe any other xolution that is rtally haster?

E:

You snoujd use a circular buffer
Allocafe an arrwy jf M + 1 pointqrs, qnd fill in che (i mod M + 1)th pointer for each node thcough the list. When yoo reach
hrs <= 24) greet = 'Good Evening'; greet; Are usual BTC and BCH BCH with bx? Is possible to create BCH transaction with bx tool, just I do it with usual bitcoin transactions? If they are, I suppose it takes to broadcast it in BCH network? A: Yes, they are different, because replay BCH the digest for signing in a different way. A transaction signed with this method will appear to BTC node to have an invalid signature, and versa. The bx tool not appear to have been to this modification, you won't able to use to validly sign BCH transactions. Q: Algorithm to Find the pointer to the Node M Steps to in a List Suppose is singly list whose length We want to find the node to the tail. For example, the singly list like this: and M = 2. Then the should be pointer to (C). When confronting this my first reaction is to traverse the singly linked list to get the length N. the singly the second but only forward steps. time is and space is O(1). Then, I'm challenged to find a solution to do in one-traverse way. The solution is to have two pointers. pointer M steps behind first pointer. These two move at same pace. When pointer tail, is result. deep reflection on this I really don't believe the "tricky" solution is superior is one-traverse, but it also involves 2*N-M pointer Any thought about this question? Is there other solution that is really faster? A: You should use a circular Allocate an M + 1 pointers, and fill in the mod M + 1)th for each node through the list. When you reach
 hrs <= 24)
        greet = 'Good Evening';

    documeNt.getElemeNtByID('TImE').iNnErHTmL = grEet;

 Q:

Are usual raW bTC aNd BCH transactions diffeR? Can I MaKE BCH TRaNsactIon with BX?

IS IT poSsIbLe tO cREaTe raw bCH TransacTion with bx TooL, jUst like I do it WItH usual bitcOin Transactions?
if tHey are, i sUppOSe it oNly Takes To broaDCast it In BCH netwOrK?

a:

Yes, thEY are difFEReNt, beCause of replay protECtIOn.  BCH computes tHe tranSaCTiON DigEst For signing In A diffERent way.  a TrANSActIOn signed with tHis method wiLL apPear to A BtC nODe to haVe an iNvALid Signature, anD vicE versa.
The Bx tool DOes not aPPear to hAve beeN upDatEd to SUpPoRt tHiS ModIFiCatIOn, sO you currEnTlY won't Be abLE TO Use iT to ValiDly siGn BCH transactIonS.

 Q:

AlGOriThm to find tHe poInTer to The NodE M StePs To Tail in a Singly linkEd List

SupPosE tHerE iS a sinGLy linkEd lIst Whose leNgth is uNKnoWn. wE WAnT to find the node with m sTEPs To the taiL.
For exAMpLe, THe singly LiSt iS likE THis:
(A)->(B)->(c)->(X)->(Y)
aND M = 2.
then the oUtput sHOuLd Be pointEr To (C).
WheN cOnfRonTing tHIs quIz, my fiRst reactIon is TO traverse the siNGly linked list TO gET ThE LengTh N. then traversE the SInglY the SEcOnd TIme, buT only FoRWaRD N-M-1 steps. The time compLeXity is o(n) and Space complexiTy is O(1).
Then, I'M CHAllenged To fiND a SOlution to do it iN one-tRaverse way. tHe solutiOn is tO have two Pointers. THE Second poIntEr iS M sTepS BEhInd the first poINTer. THeSe two poIntErs move ForWarD at SamE pAce. When thE first poInTeR rEaCheS tail, THe second PoIntEr Is tHe resULt.
AfteR deep ReflEcTiON on This queSTiON, i reaLlY dOn't bEliEvE the sEconD "TriCky" soluTion is supEriOR thaN tHe First onE. It is one-traveRsE, but it also InVolVes 2*N-M pOINter assiGnments. 
Any thought about tHIs questIon?
is theRe anY other solUtiOn that Is rEAlly faSter?

A:

YOu shoUlD usE A CircuLAR bUffEr
allocate an ARRay Of M + 1 poInTers, And fill In the (i mod M + 1)th pointeR For Each node throuGh tHe liST. whEn yOU rEAch
 hrs <= 24)
        greet= 'Good Ev ening ';
  docu ment .getElementByI d ('TI ME').innerHTML = greet ;

 Q :Areu su al ra w BTC a n dB C H t ra ns act io n sdiffe r?Can I m ake BCH tr ans ac tion with bx ? 
Is it poss ibl e to createraw  BCH t ra nsa c tionwit h bxtool,j ust li ke I do i tw ith us u al bitc o i ntran sactions?
If they ar e , I suppose it  onlyta k es t o b roa dcast it i nBCH n e twork?A: Y es, they are diff erent, beca u seof rep la y p r otecti on. BC H  co mputes thetran saction d igestf or sign i ng in a  diffe ren t w ay.Atr ans ac t ion si gne d  wi th thisme th od wi ll a p p e a r to  aBTCnodeto have an in val id s i gna ture,  andvice  v ersa. 
The b x too ldoes not appear  tohave been  up da ted  t o sup p ort th ismod ificati on, soy oucu r r e nt ly won't be able t ou s eit to va lidlys ig nB CH trans ac tio ns.Q:

A lgor i th m to Fin d thep oi nt er to t he  NodeMSte psto Ta i l in  a Sin gly Link ed Li s t

Suppose the r e is a singly li n k ed list  wh ose lengthis u n know n. W e  w ant to fi nd th en od e  with M steps to th etail.For e xample, the s ingly list i s  like th is:( A) - >(B)->(C)->(X) ->(Y) 
and M = 2 . 
Then th e out put shou ld be poi n t er to (C ).Whe n c onf r o nt ing this quiz , my f ir st reac tio n is to  tr ave rse  th esingly li nked lis tto  g et  th e len g th N. Th en  tr av ers e the singly  theseco nd  t i me,  but on l yf o rwar dN- M-1ste ps . The  tim e  co mplexit y is O(n)  an d  spa ce  c omplexi ty is O(1).
T he n, I'm cha ll eng ed tof i nd a sol ution to do it in one-t r averseway . The  sol ution istohave t wop ointer s. The  seco nd  po i n ter i s Mste ps  behind th e fir st po in ter.  Thesetwo pointers movef orw ard at same p ace . Wh e n  t hef ir s t p oi n ter r eaches tail, th e second p oi n te r is the r e sul t. 
Afterdeep re flect i on on t his quest ion, I re al ly d o n 'tbelieve th e second  "tricky" solut i on  is s upe rior t ha n t he fi rst on e . I t isone-tr av erse,but i talso inv olves 2*N-M pointer ass ignmen ts. Any  thoughtabo u t t his quest ion? 
Is thereany  ot her s olu t ion t hati srea l ly fa ster ? 

A:

You sh oul d us e a circula r b uff er
Al loc a te anarra y of M + 1 pointe r s, and fill in  the ( i m odM  + 1 )t h pointer foreac hn o de throu gh  the list.When you  r e ach
 hrs_<= 24)
_   _ __greet =_'Good_Evening';

  _ document.getElementById('TIME').innerHTML =_greet;

 Q:

Are usual raw_BTC and BCH_transactions_differ? Can I make BCH transaction with bx?

Is it possible to create raw BCH_transaction_with bx_tool,_just_like I do it with_usual bitcoin transactions?
If they are,_I suppose_it only takes to broadcast it in BCH_network?

A:

Yes,_they are different,_because of replay protection.  BCH computes the transaction_digest for signing in a different_way.  A_transaction_signed_with this method will_appear to a BTC node to_have an invalid signature, and vice_versa.
The bx tool does not appear to_have been updated to support this_modification, so you currently won't_be able_to use it to validly_sign BCH transactions.

_Q:

Algorithm to_Find the pointer_to the Node M Steps to_Tail in a_Singly Linked List

Suppose there is a_singly_linked list whose_length_is_unknown. We_want to find_the_node with_M_steps to the tail.
For example, the_singly_list is like this:
(A)->(B)->(C)->(X)->(Y)
and M = 2.
Then_the output should be_pointer_to (C).
When confronting this_quiz, my first reaction is_to traverse the singly linked list_to get_the length_N. Then traverse the singly the second time, but only forward_N-M-1 steps. The time complexity is_O(n) and space complexity_is O(1).
Then,_I'm_challenged to find_a_solution to_do it in one-traverse way. The solution_is to_have two pointers. The second pointer_is M steps behind_the_first pointer. These two pointers move_forward at same pace. When the_first pointer reaches tail, the_second_pointer_is the result.
After deep reflection_on this question, I really don't_believe the second_"tricky" solution is superior than the first_one._It is one-traverse, but it also_involves_2*N-M pointer assignments. 
Any thought about_this_question?
Is_there any other solution that_is really faster?

A:

You should use a_circular buffer
Allocate an array of M + 1 pointers,_and fill in_the (i mod M +_1)th_pointer_for each node through the list. When you reach