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)

In this above solution the run time is O(3n) and we could make this as O(n) by using the dict to track the count of integers like positive, negative and zero values.

size_of_list = int(input())
list_of_int = map(int, raw_input().split(" "))

count_of_deci = {"posi" : 0,
              "neg" : 0,
              "zero" : 0}
for ele in list_of_int:
    if ele < 0:
        count_of_deci["neg"] += 1
    elif ele == 0:
        count_of_deci["zero"] += 1
    elif ele > 0:
        count_of_deci["posi"] += 1

print "%.3f" % float(float(count_of_deci["posi"])/size_of_list)
print "%.3f" % float(float(count_of_deci["neg"])/size_of_list)
print "%.3f" % float(float(count_of_deci["zero"])/size_of_list)

I did not realize you could turn list elemetns into Ints using [int(x) for x in ....], this is amazing, thank you!

oh wow, awesome! Silly Java.

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")

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")

I'm not sure if python can optimize this?
Internally I feel that this is still performing 3 loops, instead of doing only one using 3 counters.

yes

Scala!

val result = arr.groupBy(x => if (x > 0) 1 else (if (x<0) -1 else 0)).mapValues(y => y.size.toDouble / arr.size)

println(result.getOrElse( 1, 0)) // positive
println(result.getOrElse(-1, 0)) // negative
println(result.getOrElse( 0, 0)) // zeroes

Its no where near draw :)

Here is a PHP version. $arr is the input array.

`$res = array_reduce($`arr, function(`$carry, $`val) {(`$val>0)?$`carry[0]++:((`$val<0)?$`carry[1]++:`$carry[2]++); return $`carry;},[]);    

printf ("%f\n%f\n%f", `$res[0]/$`n, `$res[1]/$`n, `$res[2]/$`n);

@paul_schmeida I am not getting my desired output as the x < 0 and x == 0 are not passing the test. I am using the Python 3 language for coding. Help me out.

three print statements with for loop, you are linearly searching 3 times the array which increases the complexity right? what special in this code other than reducing LOC

print float(len(filter(lambda x: x>0, arr)))/len(arr) print float(len(filter(lambda x: x<0, arr)))/len(arr) print float(len(filter(lambda x: x==0, arr)))/len(arr)

I think this is prettier:

n = int(input())
arr = [0 if i is 0 else i/abs(i) for i in map(int, input().split())]

for num in map(arr.count, [1, -1, 0]):
    print(round(num/n, 6))

Can you please elaborate on what "format(..)" does? I'm a beginner in python.

this may be consice but it's not very efficient because you are iterating though the array 3 times when this can be done with one iteration.

Kotlin is also really nice and clean. Especially like how clean the syntax for populating the initial array is. It has a little bit more boilerplate than Python, but still super pleasant compared to what you'd have to do in java 7 or lower.

import java.util.Scanner

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)

    val size = sc.nextInt()
    val inputs = Array(size, { sc.nextDouble() })

    println(inputs.count { it > 0.0} / size.toDouble())
    println(inputs.count { it < 0.0} / size.toDouble())
    println(inputs.count { it == 0.0} / size.toDouble())
}

why we have to use square brackets inside len()

This will not work in Python2 because you are doing integer divison.

CAn you please explain this syntax.

well first it's a cool solution but what i mean here is that in each result you are reading the whole array and this is "i think" not efficient.

Dear paul_schmeida,

can u please explain why u have used ".6f" inside the format function.Please Help.Advance thanks and regards.

Just so we're clear, you're aware there are three different loops, right? Seems pretty inefficient to me.

def plusMinus(arr): lis1=[x for x in arr if x>0] lis2=[x for x in arr if x<0] lis3=[x for x in arr if x==0] print(f'{round(len(lis1)/len(arr),6)}\n{round(len(lis2)/len(arr),6)}\n{round(len(lis3)/len(arr),6)}')

Going three times through the input data?

Am I reading this right that this will iterate over the array three times? If thats the case while this looks nice, its less efficient!

Your code is extremely inefficient, it only looks short. You are using three loops when you require only 1. The complexity of your solution is O(3n)

print('%.6f'%(len(list(filter(lambda num: num>0, arr)))/len(arr)))

print('%.6f'%(len(list(filter(lambda num: num<0, arr)))/len(arr)))

print('%.6f'%(len(list(filter(lambda num: num==0, arr)))/len(arr)))

time complexity goes crazy, watch out!