HackerRank ‘Plus Minus’ Solution

Short Problem Definition:

You’re given an array containing integer values. You need to print the fraction of count of positive numbers, negative numbers and zeroes to the total numbers. Print the value of the fractions correct to 3 decimal places.


Plus Minus


time complexity is O(N)

space complexity is O(1)


Count the values. Do not forget to force floats instead of integers.


def getPartitionPythonesque(values):
    c1 = len(filter(lambda x:x>0,values))
    c2 = len(filter(lambda x:x<0,values))
    c3 = len(filter(lambda x:x==0,values))
    v_len = float(len(values))
    return (c1/v_len, c2/v_len, c3/v_len)

def getPartition(values):
    pos, neg, zero = [0.0,0.0,0.0]
    v_len = len(values)
    for value in values:
        if value == 0:  zero += 1
        elif value > 0: pos += 1
        else:           neg += 1
    return (pos/v_len, neg/v_len, zero/v_len)    

if __name__ == '__main__':
    t = input()
    values = map(int, raw_input().split())
    partition = getPartition(values)
    for percentage in partition:
        print round(percentage,4)

use std::io;
use std::cmp::Ordering;

fn get_number() -> u32 {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");

fn get_numbers() -> Vec<i32> {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::<i32>().unwrap()).collect()

fn main() {
    let siz = get_number() as f32;
    let (mut pos, mut neg, mut zer) = (0, 0, 0);
    let zero = 0;
    for i in get_numbers() {
        match i.cmp(&zero) {
            Ordering::Less => neg+=1,
            Ordering::Greater => pos+=1,
            Ordering::Equal => zer+=1,
    println!("{:.6}\n{:.6}\n{:.6}", pos as f32 /siz, neg as f32 /siz, zer as f32 /siz);

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  • Stephen Estrada

    As far as Python 3.* you can’t call len on filter, this will raise a TypeError: object of type ‘filter’ has no len().
    You can, however, do this: v1 = list(filter(lambda x: x > 0, values)) and call len(v1), using list(), otherwise you get a . I am, however, uncertain of the time complexity of this problem. Either way this solution will work. Thought I’d let you know!

    • It’s a three pass algorithm where only one pass is necessary. I strongly prefer the other version I posted.

      I have to check filters is 3.* I am still with 2.7.* :). Thanks for sharing this information