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:
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 space-separated integers that describe arr[n]
Constraints
0 < n < 100
-100 <= arr[i] <=100
Output Format
Print the following
lines, each to
decimals:
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++)
{
if(arr[i]>0)
p_count += 1;
else if (arr[i]<0) {
n_count +=1;
}
else if(arr[i]==0) z_count += 1;
}
printf("%.5f\n",(p_count/arr_count));
printf("%.5f\n",(n_count/arr_count));
printf("%.5f\n",(z_count/arr_count));
}
In Java :
import java.io.*;
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(System.in);
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) {
pos++;
} else if (x == 0) {
zero++;
} else {
neg++;
}
}
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;
cin>>j;
for(i=0;i<j;i++){
cin>>a;
if(a>0)
p++;
else if(a<0)
n++;
else
z++;
}
printf("%.3f\n",(float)p/j);
printf("%.3f\n",(float)n/j);
printf("%.3f",(float)z/j);
/* 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
else:
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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