plusMinus.js

// 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  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  are acceptable.

// Example

// There are  elements, two positive, two negative and one zero. Their ratios are ,  and . Results are printed as:

// 0.400000
// 0.400000
// 0.200000
// Function Description

// Complete the plusMinus function in the editor below.

// plusMinus has the following parameter(s):

// int arr[n]: an array of integers
// Print
// Print the ratios of positive, negative and zero values in the array. Each value should be printed on a separate line with  digits after the decimal. The function should not return a value.

// Input Format

// The first line contains an integer, , the size of the array.
// The second line contains  space-separated integers that describe .

// Constraints



// Output Format

// Print the following  lines, each to  decimals:

// proportion of positive values
// proportion of negative values
// proportion of zeros
// Sample Input

// STDIN           Function
// -----           --------
// 6               arr[] size n = 6
// -4 3 -9 0 4 1   arr = [-4, 3, -9, 0, 4, 1]
// Sample Output

// 0.500000
// 0.333333
// 0.166667
// Explanation

// There are  positive numbers,  negative numbers, and  zero in the array.
// The proportions of occurrence are positive: , negative:  and zeros: .



// SOLUTION


function plusMinus(arr) {
    // Write your code here
    var len = arr.length
    let neg = 0
    let pos = 0
    let zero = 0
    for (var i = 0; i < len; i++) {
        if (arr[i] > 0) {
            pos++
        }
        else if (arr[i] < 0) {
            neg++
        }
        else {
            zero++
        }
    }
    console.log((pos / len).toPrecision(6))
    console.log((neg / len).toPrecision(6))
    console.log((zero / len).toPrecision(6))

}