theroadtaken123 You can solve this easily using a bit of math:
y=int(raw_input()) z=y while(z%3!=0): z-=5 if(z<0): print '-1' else:
print z*'5'+(y-z)*'3'If the number(say 66317) is not divisible by 3, it will leave a modulo of either 0,1 or 2. If I decrease the number by 5, I am basically making it a multiple of 3, and the remaning digits will be a multiple of 5 as I am subtracting it from the number.
modulo 0 implies number divisibile modulo 1 implies 5 needs to be subtracted twice. modulo 2 implies 5 needs to be subtracted once.
Correct me if I am wrong. Completed all the test cases in 0-0.01s
snapdragon1 [deleted]vishalbty Here's how I solved it using divmod.
yeremy You're my hero. Thanks for the tip!!
theroadtaken123 Thank you yeremy, appreciate the reply :)
Amudala Thanks for posting it.
pardonmydust4 [deleted]
Mang_ It's a great solution. If I understand correctly, you separate N in two numbers: a multiple of 3 and a multiple of 5.
Is there a name to this mathematical procedure ?
rosiezou It is called a linear combination. In this case, it is a linear combination of 3 and 5.
edz65536 More specifically it is a Diophantine equation.
amdokamal There is an alternative solution, but not so efficient (but still O(1) in spite of recursion) and not so elegant as the Diophantine equation. This is linear congruence equation: 5x = n (mod 3). You can solve it by means of the extended Euclidean algorithm. I am aware that this is like 'A sledgehammer to crack a nut', but fyi. Here you can practise with the linear congruence.
theroadtaken123 Thank you mang, appreciate the reply :)
AntonDeCode Pro Tip for Java user: Use the StringBuilder class.
String is immutable, so everytime you add to a string, it creates a new object! With an input of 100000, you'd have created 30000 new objects each containing the string it was before it plus what you added to it! You can imagine how innefficient this is.
StringBuilder, 1 object, 1 problem solved. Use StringBuilder today!
(This message is brought to you by StringBuilder Co. We build Strings in a timely manner for you, so you don't have to!)
ifigu003 Thanks man. Now all the test cases are accepted.
yeremy Wow... didn't know how powerful StringBuilder is. Thank you!
pardonmydust4 ok you are the best
Hmpflabul MVP :')
asbear This function works quicker than of editorial.
// return negative number if not found // otherwise pivot is returned. // caller can print 5 x pivot + 3 x (n-pivot) int getPivot(int n) { while(n > 0) { if(n%3 == 0) break; else n -= 5; } return n; }So you can do this:
int pivot = getPivot(n); if(pivot < 0) cout << "-1" << endl; else { int repeat = pivot/3; while(repeat--) cout << "555"; repeat = (n-pivot)/5; while(repeat--) cout << "33333"; cout << endl; }edmondkwan am i missing somethign here? The editorial solution is O(1) how is this faster?
BSathvik you see O(1) means that its constant time but the constant(Eg:O(1000000)=O(1)) however big is rewritten as O(1) for the sake of simplicity. well maybe what he is trying to say is that maybe his constant is smaller(i am not sure if it is). i i am just trying to tell you that Alg1 maybe faster than Alg2 even though both of them hve the complexity of O(1).
SnoopDizzle The loop in this iterative approach will run at most 2 times, which technically makes this solution O(1) as well. However, it should be obvious that the direct formula will require fewer machine instructions. By counting just the math operations (so ignoring jumps, which inflate the cost even more), it breaks down like this (please comment if I made a mistake!):
Direct (2 x N % 3): + 1 Multiply + 1 Mod = 2 ConstantlyIterative:
Best Case (N is divisible by 3): + 1 Comparison ( while(n > 0) ) + 2 Comparison and Mod ( if(n%3 == 0) ) = 3 Worse Case (N % 3 == 1), full loop runs twice: + 3 Comparison ( while(n>0) ) + 6(3x2) Comparison and Mod ( if(n%3 == 0) ) + 2(2x1) Subtraction ( n -= 5 ) = 11And we might as well do the 3rd case (loop runs once):
+ 2 Comparison ( while(n>0) ) + 4(2x2) Comparison and Mod ( if(n%3 == 0) ) + 1 Subtraction ( n -= 5 ) = 7Since each case is equally likely, we calculate that the average cost is (3+7+11)/3 = 7 instructions.
So although both solutions are O(1), we can correctly conclude that the iterative approach is ~3 times more expensive. If we counted jumps and assignments, it's more like ~4 times. This is a significant difference.
big_endian It fails in some cases for ex: when N=1000 :P. PS: adding if(pivot%3!=0) pivot=getPivot(pivot); before checking for if(pivot<0) solves the prolem.
asbear I might be wrong :). But I can still get a result even with N=1000. What is the correct answer for N=1000? Also I don't understand why you need to check (pivot%3 != 0) while it is checked in getPivot() function.
taitung My answer to N = 1000 is 5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555553333333333
you see there are 10 number 3 in this string.
sagemode yeah finding the pivot is important. My solution is similar. I aimed to remove the expensive operation "modulo", here's my solution.
#Python solution without modulo t = int(input().strip()) for a0 in range(t): n = int(input().strip()) x = (int(n / 3)) * 3 y = 0 while True: y = n - x if ((int(y / 5)) * 5 != y): if (x == 0): print("-1") break x = (int((x - 1) / 3)) * 3 else: print (int(x) * "5" + int(y) * "3") break
mdsmithson why are you putting a solution in the discussion?
gauraviitkgp [deleted]
gauraviitkgp [deleted]gauraviitkgp import sys can u telll me what is error in my code
t = int(raw_input())
test = 0
for a0 in xrange(t):
n = int(raw_input().strip()) for i in range(0,n): if (n-5*i)>=0 and (n-5*i)%3==0: print("5"*(n-5*i)+"3"*(i*5)) test = test + 1 if test==0: print("-1")
flowerking Downloaded and checked the test cases on local machine. And the answers correct but when submitted showing as incorrect!
thebick Apparently HackerRank is truncating all output after 32767 total characters (on all lines combined). Perhaps someone needs to check the output box to see what its limit is.
CattleOfRa my code consists of 600 characters, it's working perfectly fine on my local machine but gives me timeout here. What could be the issue?
CattleOfRa Nevermind. My algorithm was just too slow. I've managed to complete it now
changcw83 [deleted]dexxer I'm running into the same issue. Downloaded a few of the failed test cases only to find out that my algorithm actually works... lame!
wasihaiderdev Probably a performance issue...
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