wow. thats pretty smart can you tell me where you learned it or walk me through your though proccess for

int *A = new int[N];

Can you explain me this step: " while(cout<

Heres another solution:

int main() {
    int n;  
    cin >> n;
    const int a = n;
    int arr[a];

    return 0;
}

(jorgechiquinv) Yep! Same here (excpet without the whiles, that looks clean, I like it). I threw a couple of runtime exceptions to handle the constraints on N and Ai. I also had N as a reference to a variable named arraySize just for fun :)

Why did you add '&& N' in 'while(cout<

thanks, Jorge, that worked for me

what 's "and" in while mean?

while(cout<

delete[] A; what's this statement doing?

First of all thanks for sharing your solution. I really appreciate it. But i found a bug. There is no exit condition for the first while loop. It will keep writing in the unauthorised memory and crash. It should be like this.

int main()
{
    int N,i=0;
    cin>>N;
    int *A = new int[N];
    while(cin>>A[i++] && (i < N));
    while(cout<<A[--N]<<' ' && N);
    delete[] A;
    return 0;
}

there should be one more condition in the first while loop

while(cin>>A[i++] and i<N);

otherwise, will be an infinit loop. thumps up, nice approach

can u pls explain me both the while statements, i am not getting it.

Can you explain, how did the array get reversed?

Hi, the snippet is not working in codeblocks. Is there any particular reason behind it?

P.s - I am new to this.

Truly elegant. Shouldn't we free A or set it to NULL as well?

why we must cout N

For those of you who are wondering how the 1st while loop will terminate , it will terminate only if you type 'false' or 'eof' . It is a fancy way of solving this question but it is not optimised way because it takes a hell lot of machine cycles and instructions . take a look

while (cin >> A[i++] && (i < N)); 00B324BD mov eax,dword ptr [i]
00B324C0 mov ecx,dword ptr [A]
00B324C3 lea edx,[ecx+eax*4]
00B324C6 mov dword ptr [ebp-11Ch],edx
00B324CC mov eax,dword ptr [i]
00B324CF add eax,1
00B324D2 mov dword ptr [i],eax
00B324D5 mov esi,esp
00B324D7 mov ecx,dword ptr [ebp-11Ch]
00B324DD push ecx
00B324DE mov ecx,dword ptr [imp?cin@std@@3V?char_traits@D@std@@@1@A (0B3D098h)]
00B324E4 call dword ptr [__imp_std::basic_istream >::operator>> (0B3D09Ch)]
00B324EA cmp esi,esp
00B324EC call __RTC_CheckEsp (0B312DFh)
00B324F1 mov dword ptr [ebp-0FCh],eax
00B324F7 mov edx,dword ptr [ebp-0FCh]
00B324FD mov eax,dword ptr [edx]
00B324FF mov ecx,dword ptr [ebp-0FCh]
00B32505 add ecx,dword ptr [eax+4]
00B32508 mov esi,esp
00B3250A call dword ptr [__imp_std::ios_base::operator bool (0B3D0D0h)]
00B32510 cmp esi,esp
00B32512 call __RTC_CheckEsp (0B312DFh)
00B32517 movzx ecx,al
00B3251A test ecx,ecx
00B3251C je main+0E8h (0B32528h)
00B3251E mov eax,dword ptr [i]
00B32521 cmp eax,dword ptr [N]
00B32524 jge main+0E8h (0B32528h)
00B32526 jmp main+7Dh (0B324BDh)
while (cout << A[--N] << ' ' && N); 00B32528 mov eax,dword ptr [N]
00B3252B sub eax,1
00B3252E mov dword ptr [N],eax
00B32531 push 20h
00B32533 mov esi,esp
00B32535 mov ecx,dword ptr [N]
00B32538 mov edx,dword ptr [A]
00B3253B mov eax,dword ptr [edx+ecx*4]
00B3253E push eax
00B3253F mov ecx,dword ptr [imp?cout@std@@3V?char_traits@D@std@@@1@A (0B3D0D4h)]
00B32545 call dword ptr [__imp_std::basic_ostream >::operator<< (0B3D0A4h)]
00B3254B cmp esi,esp
00B3254D call __RTC_CheckEsp (0B312DFh)
00B32552 push eax
00B32553 call std::operator<< > (0B314A6h)
00B32558 add esp,8
while (cout << A[--N] << ' ' && N); 00B3255B mov dword ptr [ebp-108h],eax
00B32561 mov ecx,dword ptr [ebp-108h]
00B32567 mov edx,dword ptr [ecx]
00B32569 mov ecx,dword ptr [ebp-108h]
00B3256F add ecx,dword ptr [edx+4]
00B32572 mov esi,esp
00B32574 call dword ptr [__imp_std::ios_base::operator bool (0B3D0D0h)]
00B3257A cmp esi,esp
00B3257C call __RTC_CheckEsp (0B312DFh)
00B32581 movzx eax,al
00B32584 test eax,eax
00B32586 je main+150h (0B32590h)
00B32588 cmp dword ptr [N],0
00B3258C je main+150h (0B32590h)
00B3258E jmp main+0E8h (0B32528h)

this is nice but allocating memory havent been tocuched in previous lessons...

thanks bro, it was helpful.

your code requires to enter char/string so it can print the value

Despite the code posted by JChiquin passed all of the test cases, the Visual Studio console did not give the same output until I modified the first while loop as:

while (std::cin >> A[i++] && i<N);

Did anyone stop by the same issue?

You may reuse N and ditch i

that helped a lott thanks very much

Some body please explain me both while loops like how they are working and how it is giving us the desired result.

Program won't run as required. it will take input till infinite. you used new here when it crosses the N length. Itself it will double the array size and so on.

Here you did not give the break loop condition for the first while loop. so it will take input till N and store it in array of size N after that it will again take the value and put it in buffer which no means. and if u enter any charachter then it will exit from the loop because here type is integer. break condition: while(std::cin>>A[i++] && i

How will you check if

while(std::cin>>A[i++]); talking more input than N?

I mean when this loop will end?

awesome solutuion bro!

i'd love your way and it's passed on hackarank but i tested it on visual studio, the input of N is a loop without break. i have modified a little to:

include

int main() { int N, i = 0; std::cin >> N; int* A = new int[N]; while (i> A[i++]); while (std::cout << A[--N] << ' ' && N); delete[] A; return 0; }

hello , i undestood you have dynamically created array ans assigned the input as size but can you explain below two lines: while(std::cin>>A[i++]); while(std::cout<

will not this line -- while(std::cin>>A[i++]); ---- loop indefinitely ? solution can be -- while(std::cin>>A[i++] &&(i!=N)); ---

no exit condition written for first while loop ????

Hey I'm just a bit confused. Using the example input, 4 and 1 4 3 2, when you say int numElements, numElements = 4 right? And doing int arr[numElemnts] that creates an array that includes 0 through 3?

in input section when i run in c++ it isnt leaving space

I get the concept but for some reason I don't understand it very well (this question I mean). Can you please put comments in your code to explain properly what's happening here? I would really appreciate it... thanks.

if i start loop i=o then compiler display error.when i=1 then ok but again show compiler error because quetion demand i/o using stdin and stdout so how i face difficulty ,please give me solution

Hey I am using visual studio and apparently int arr[numElements] wouldn't work (error of "expression must have a constant value")

In first for loop if you replace cin >> arr[i] with following: int x; cin >> x; arr[i] = x;

That will work otherwise it will fail if you leave your code as it is and try following values it will fail:

8 6676 3216 4063 8373 423 586 8850 6762

mine is also same but its showing some test case failed and compiler error occure as segmentation error.........

I did the same thing, but some of the testcases get abort called! Is this because some of the numbers are too big?

i too figured out th same logic but i can not see the output on screen ???

thanks. it worked. but I still am not able to find why my program showed "Segmentation fault" when I wrote the same exact code.

for (int i = numElements - 1; i >= 0; i--) cout << arr[i] << " "; can you plesae explain this loop's conditions

but when i do like this ..im getting runtime errors :(

Nice, learned that I can define the initial i in the for loop.

Once my teachers said me that declaring a pointer like int numElements; cin >> numElements; int arr[numElements]; would lead me to static memory issues. They suggest me to implement dinamic memory. Is there a special fact to ignore my teachers in this case? Help me, because i don't see anyone complained with this. Sorry for my english.

Its saying Segmentation fault. I also came up with the same solution. Did this problem occur with you as well?

i also think the same logic........

why is it for (int i = numElements - 1; i >= 0; i--) instead of for (int i = numElements - 1; i = 0; i--)?

Can anyone explain this to me? Thank you!

why a-1 ????? in second for loop

I did the variable array length solution, and tried compiling it with clang++, that worked fine as well.

Thanks for clarifying. I had thought that specifying the size of an array at runtime was not allowed in C++, but then I was second-guessing myself because the code compiled and ran fine.

Thank you for explaining this. I was very confused as to why the compiler let me get away with declaring an array size via a variable, instead of forcing me to make a call to new.

Dunno why those but in fact vector is much more heavier than plain array. So there is a good point in using it instead vector. Imho.

Does size_t size{} == size_t size ? what does the {} means here please?

Using g++ I get compile errors on this one.

Hey Ethan, I tried following this approach. I got these errors: Reading symbols from Solution...done New LWP 163882 Thread debugging using libthread_db enabled Using host libthread_db library "/lib/x86_64-linux-gnu/libthread_db.so.1 Core was generated by `./Solution' Program terminated with signal SIGSEGV, Segmentation fault

0 0x00007faa5e714e20 in std::istream::operator>>(int&) ()

from /usr/local/lib64/libstdc++.so.6 To enable execution of this file add and few other errors Can someone help??

easy solution superb