Algorithm
Problem Name: Algorithms -
In this HackerRank Functions in Algorithms - Java programming problem solution,
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 power -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:
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 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 n space-separated integers that describe arr[n]
Constraints
0 <= n <= 100
-100 <= arr <= 100
Output Format
Print the following 3 lines, each to 6 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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main() {
int n, k;
float pos = 0, neg = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &k);
if (k < 0) {
neg++;
} else if (k > 0) {
pos++;
}
}
printf("%f\n%f\n%f", pos / n, neg / n, (n - pos - neg) / n);
return 0;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int p=0;int m=0;int z=0;
int main()
{ int n;
cin>>n;
int ar[n];
for(int i = 0; i < n; i++)
{
cin >> ar[i];
}
for(int i = 0; i < n; i++)
{ if(ar[i] > 0)
p++;
else if(ar[i] < 0)
m++;
else z++;
}
double a1,a2,a3;
a1 = double(p)/double(n);a2 = double(m)/double(n);a3 = double(z)/double(n>;
cout << a1 << "\n" << a2 << "\n" << a3;
return 0;
}
#3 Code Example with Java Programming
Code -
Java Programming
import java.util.Scanner;
public class averageratio {
// Java program to find the ratio of positive,
// negative, and zero elements in the array.
// Function to find the ratio of
// positive, negative, and zero elements
public static void positiveNegativeZero(int[] arr)
{
// Store the array length into the variable len.
int len = arr.length;
// Initialize the postiveCount, negativeCount, and
// zeroCountby 0 which will count the total number
// of positive, negative and zero elements
float positiveCount = 0;
float negativeCount = 0;
float zeroCount = 0;
// Traverse the array and count the total number of
// positive, negative, and zero elements.
for (int i = 0; i < len; i++) {
if (arr[i] > 0) {
positiveCount++;
}
else if (arr[i] < 0) {
negativeCount++;
}
else if (arr[i] == 0) {
zeroCount++;
}
}
// Print the ratio of positive,
// negative, and zero elements
// in the array up to four decimal places.
System.out.printf("%1.6f ", positiveCount / len);
System.out.println();
System.out.printf("%1.6f ", negativeCount / len);
System.out.println();
System.out.printf("%1.6f ", zeroCount / len);
System.out.println();
}
// Driver Code.
public static void main(String args[])
{
// System.out.println("Enter the elements size for Ist test case");
Scanner sc = new Scanner(System.in);
int n= sc.nextInt();
// Test Case 1:
int a1[]= new int[n] ;
for (int i = 0; i < n; i++){
a1[i] = sc.nextInt();
}
// int[] a1 = { 2, -1, 5, 6, 0 };
positiveNegativeZero(a1);
// Test Case 2:
/* int[] a2 = { 4, 0, -2, -9, -7, 1 };
positiveNegativeZero(a2); */
}
}
#4 Code Example with C# Programming
Code -
C# Programming
using System.CodeDom.Compiler;
using System.Collections.Generic;
using System.Collections;
using System.ComponentModel;
using System.Diagnostics.CodeAnalysis;
using System.Globalization;
using System.IO;
using System.Linq;
using System.Reflection;
using System.Runtime.Serialization;
using System.Text.RegularExpressions;
using System.Text;
using System;
class Solution {
static void plusMinus(int[] arr) {
int n = arr.Length;
int positives = arr.Where(x => x > 0).Count();
int negatives = arr.Where(x => x < 0).Count();
int zeroes = arr.Where(x => x == 0).Count();
float positiveFraction = (float)positives / (float)n;
float negativeFraction = (float)negatives / (float)n;
float zeroFraction = (float)zeroes / (float)n;
Console.WriteLine(positiveFraction);
Console.WriteLine(negativeFraction);
Console.WriteLine(zeroFraction);
}
static void Main(string[] args) {
int n = Convert.ToInt32(Console.ReadLine());
int[] arr = Array.ConvertAll(Console.ReadLine().Split(' '), arrTemp => Convert.ToInt32(arrTemp));
plusMinus(arr);
}
}
#5 Code Example with Python Programming
Code -
Python Programming
import math
import os
import random
import re
import sys
# Complete the plusMinus function below.
def plusMinus(arr):
pos=0
neg=0
neu=0
for i in range(len(arr)):
if(arr[i]>0):
pos+=1
elif(arr[i]==0):
neu+=1
else:
neg+=1
print(pos/len(arr))
print(neg/len(arr))
print(neu/len(arr))
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().rstrip().split()))
plusMinus(arr)