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I felt like cheating too so opened the discussion forum for the first time to see this post :D
n=float(raw_input())
numbers=map(int,raw_input().split())
print round(len([x for x in numbers if x>0])/n,3)
print round(len([x for x in numbers if x<0])/n,3)
print round(len([x for x in numbers if x==0])/n,3)
n = float(raw_input())
lst = [int(x) for x in raw_input().split()]
print format(len([x for x in lst if x > 0])/n, ".6f")
print format(len([x for x in lst if x < 0])/n, ".6f")
print format(len([x for x in lst if x == 0])/n, ".6f")
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
double a=0,b=0,c=0;
int arr[] = new int[n];
for(int arr_i=0; arr_i < n; arr_i++){
arr[arr_i] = in.nextInt();
}
for(int arr_i=0; arr_i < n; arr_i++){
if (arr[arr_i]>0)
{a++;}
else if (arr[arr_i]<0){b++;}
else if (arr[arr_i]==0){c++;}
}
System.out.println(a/n);
System.out.println(b/n);
System.out.println(c/n);
}
Could have done the conditionals and assigning arr the scanner.nextInt() numbers all in one for loop:
importjava.io.*;importjava.util.*;publicclassSolution{publicstaticvoidmain(String[]args){/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */Scannerscanner=newScanner(System.in);intsize=scanner.nextInt();int[]a=newint[size];doublefractionPositive=0;doublefractionZero=0;doublefractionNegative=0;for(inti=0;i<size;i++){a[i]=scanner.nextInt();if(a[i]>0)fractionPositive+=1;elseif(a[i]==0)fractionZero+=1;elseif(a[i]<0)fractionNegative+=1;}System.out.println(fractionPositive/size);System.out.println(fractionNegative/size);System.out.println(fractionZero/size);}}
You also don't need to count one of them, for instance, if you don't count negative ones, your output is:
System.out.println((size-fractionPositive-fractionZero)/size);
Most programing languages do not like when variable types mix with eachother. An integer divided by an integer produces a floating point in most cases, but the computer only works with variables that all have the same data type; such as a double divided by a double producing a double. The problem seems like you are having is because of inconsistant varible types.
The folowing case does not work because there are two differant variable types:
double = integer / integer
A solution to this problem is to use a STATIC CAST. The following line could be read as "turn my (positiveIntegerCount) into a double before i divide by its length":
Because, in Python, if you divide two ints, you would get the result of the integer division, in this case 0. But if the N is stored as a float, you are already forcing the floating point division here.
int n = Convert.ToInt32(Console.ReadLine());
string[] arr_temp = Console.ReadLine().Split(' ');
int[] array_temp= Array.ConvertAll(arr_temp,Int32.Parse);
int PositiveNumbers= array_temp.Where(x=>x>0).ToList().Count;
int NegativeNubers=array_temp.Where(x=>x<0).ToList().Count;
int zeroNumbers=array_temp.Length-(PositiveNumbers + NegativeNubers);
head = float(raw_input())
mylist = [float(i) for i in raw_input().split()]
zero = sum(j == 0 for j in mylist)
ng = sum(j < 0 for j in mylist)
ps = sum(j > 0 for j in mylist)
print round(ps/head, 3)
print round(ng/head, 3)
print round(zero/head, 3)
When i read it correctly, she walks through 'mylist' 3 times. If she would go through each element once and do the compareision in that loop, she would walk through the list only once.
But because she also reads the input in a list and than splits it, she basicly is creating a temp list which isn't necessary. In java you can say (when using a Scanner) nextInt and don't even need a list at all.
As far as complexity goes, repeating an iteration 3 times is the same as doing it once so it's not the end of the world. Especially with the constraint that N <100.
The code idea is correct but we don't need to perform three O(n) operations for same stuff. The time complexity will remain O(n) but if taken for large array length, there will be slight difference in time in microseconds probably compared to that with one or two O(n) time complexity program.
We mostly don't care for space complexity (just for single list), thus, we may use it.
Guys! I have a question related to Python, in this piece of code of yours - [x for x in numbers if x>0].
What if I wanted to create a "new list" (which contains only negative numbers) from "numbers list" where numbers list is sorted in ascending order. I want to break the loop when x>= 0. So,
a = [x for x in numbers if x >= 0: break]
As you know this syntax is wrong, can you guys suggest me what is the proper way? I am new to Python.
Plus Minus
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I felt like cheating too so opened the discussion forum for the first time to see this post :D
High five!
Quick help? How do I format code snippet in my reply? I used
and it isn't formatted. Thanks in advance.Nevermind! got it from a comment below :)
n = float(raw_input()) lst = [int(x) for x in raw_input().split()] print format(len([x for x in lst if x > 0])/n, ".6f") print format(len([x for x in lst if x < 0])/n, ".6f") print format(len([x for x in lst if x == 0])/n, ".6f")
public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); double a=0,b=0,c=0; int arr[] = new int[n]; for(int arr_i=0; arr_i < n; arr_i++){ arr[arr_i] = in.nextInt(); }
Could have done the conditionals and assigning arr the scanner.nextInt() numbers all in one for loop:
Should've used printf for more precise result. Cool solution tho'.
https://www.hackerrank.com/challenges/plus-minus/forum/comments/236802
is it ("%.6f % n") regular expression and how it is formatting the result
Ahh got it to limit the output after decimal..
but in between " " this will coment than how wil be this work??
thank u...
("%.6f %n", this "," sould we use +
no this will print the whole string
import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
} why this code is not being approved???
because there is one \n at the beginning which is not necasary
System.out.printf("%1$.6f", p/al);
add \n after 6f in each s.o.pln then it will pass all test cases
just remove \n from first printf not add !
just remove \n from first printf as its starting from end of line which is against the format
cool solution bro
We can also use float instead of double. It will work.
Didn't really need an array.
You also don't need to count one of them, for instance, if you don't count negative ones, your output is: System.out.println((size-fractionPositive-fractionZero)/size);
This could be used as a replacement for your print statement
good solution to union all sout.
Can someone exaplin how to get this regex? %.06f%n%.06f%n%.06f I am a little confused
It is not a regex, Google pls - print format in Java
awesome thanks
i am trying to do the same thing but the output remains 0.000000 i dont know how?
Are your variables of type
doubleinstead ofint?no.
Same for me...
yea.. same for me also...
Most programing languages do not like when variable types mix with eachother. An integer divided by an integer produces a floating point in most cases, but the computer only works with variables that all have the same data type; such as a double divided by a double producing a double. The problem seems like you are having is because of inconsistant varible types.
The folowing case does not work because there are two differant variable types: double = integer / integer
A solution to this problem is to use a STATIC CAST. The following line could be read as "turn my (positiveIntegerCount) into a double before i divide by its length":
typecast into float
same here....!
apply casting while dividing number of positive(negative or zeroes)by number of elements e.g. a=(float)b/n;
This won't give decimal output
but what is data type of a&b&c???
Why your "N" a float , it is the length of the Array and i think it must be integer because you can't an Array with 2.5 in size !!!
Because, in Python, if you divide two ints, you would get the result of the integer division, in this case 0. But if the N is stored as a float, you are already forcing the floating point division here.
To my knowledge, that is only the case for Python 2. Python 3 returns the fractional part as well even if you divide two ints. :)
yeah I think that's right
Right, in python3 if you dont want the fraction just use floor division a//x.
In C# as well. Took me "forever" to find out.
int n = Convert.ToInt32(Console.ReadLine()); string[] arr_temp = Console.ReadLine().Split(' '); int[] array_temp= Array.ConvertAll(arr_temp,Int32.Parse); int PositiveNumbers= array_temp.Where(x=>x>0).ToList().Count; int NegativeNubers=array_temp.Where(x=>x<0).ToList().Count; int zeroNumbers=array_temp.Length-(PositiveNumbers + NegativeNubers);
you could also import division from future which fixes the integer and float division as in Python 3.
nice idea! not wasting memory and more importantly time in creating a list. Thanks for sharing!
When i read it correctly, she walks through 'mylist' 3 times. If she would go through each element once and do the compareision in that loop, she would walk through the list only once.
But because she also reads the input in a list and than splits it, she basicly is creating a temp list which isn't necessary. In java you can say (when using a Scanner) nextInt and don't even need a list at all.
As far as complexity goes, repeating an iteration 3 times is the same as doing it once so it's not the end of the world. Especially with the constraint that N <100.
The code idea is correct but we don't need to perform three O(n) operations for same stuff. The time complexity will remain O(n) but if taken for large array length, there will be slight difference in time in microseconds probably compared to that with one or two O(n) time complexity program. We mostly don't care for space complexity (just for single list), thus, we may use it.
REDUCE COMPLEXITY TO WHATEVER YOU CAN!!
can you please explain this? scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
Nice ! But I think its better to use format than round since round will not display zeros after decimal.
Try this:
can you plzz explain this scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
Guys! I have a question related to Python, in this piece of code of yours - [x for x in numbers if x>0].
What if I wanted to create a "new list" (which contains only negative numbers) from "numbers list" where numbers list is sorted in ascending order. I want to break the loop when x>= 0. So,
As you know this syntax is wrong, can you guys suggest me what is the proper way? I am new to Python.
There is no recommended way to interrupt an "expression" loop (i.e., comprehension, filter, map, reduce). You must use a conventional loop.
More on this, with a hack: http://stackoverflow.com/questions/9572833/break-list-comprehension
Depends if you really need to break.
If not -- simply:
why u r using n as float,how no. of elements can be float
please explain me the print part
Its less code, but less efficent than just iterating the list once isnt it?