Plus Minus

Problem Statement :

Given an array of integers, calculate the ratios of its elements that are positive, negative, and zero. Print the decimal value of each fraction on a new line with 6 places after the decimal.

Note: This challenge introduces precision problems. The test cases are scaled to six decimal places, though answers with absolute error of up to 10^ -4 are acceptable.

Example :
arr = [1,1,0, -1,-1]

There are n =5 elements, two positive, two negative and one zero. Their ratios are 2/5=0.400000 , 2/5=0.400000  and 1/5 = 0.200000 . Results are printed as: 


Function Description

Complete the plusMinus function in the editor below.

plusMinus has the following parameter(s):

    int arr[n]: an array of integers

Print the ratios of positive, negative and zero values in the array. Each value should be printed on a separate line with 6 digits after the decimal. The function should not return a value.

Input Format

The first line contains an integer n, the size of the array.
The second line contains space-separated integers that describe arr[n]


0 < n < 100
-100 <= arr[i] <=100

Output Format

Print the following
lines, each to


1. proportion of positive values
2. proportion of negative values
3. proportion of zeros

Solution :


                            Solution in C :

In C :

void plusMinus(int arr_count, int* arr) {

    float p_count = 0, n_count = 0 , z_count = 0;
    for(int i = 0; i< arr_count; i++)
        p_count += 1;

        else if (arr[i]<0) {
        n_count +=1;

        else if(arr[i]==0) z_count += 1;



In Java : 

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(;
        int n = in.nextInt();
        int pos = 0;
        int zero = 0;
        int neg = 0;
        for (int i = 0; i < n; i++) {
            int x = in.nextInt();
            if (x > 0) {
            } else if (x == 0) {
            } else {
        System.out.println(pos / (double) n);
        System.out.println(neg / (double) n);
        System.out.println(zero / (double) n);

In C ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int p=0,n=0,z=0,a,i,j;
        else if(a<0)
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;

In Python3 :

N = int(input())
listahan = input().split()
diks = {"pos": 0, "neg": 0, "zer": 0}
for i in listahan:
    if int(i) > 0:
        diks["pos"] += 1
    elif int(i) < 0:
        diks["neg"] += 1
        diks["zer"] += 1     
print(format(diks["pos"]/N, '.3f'))
print(format(diks["neg"]/N, '.3f'))
print(format(diks["zer"]/N, '.3f'))

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