How To Solve The Plus Minus Code Challenge [C programming language]

How To Solve HackerRank’s Plus Minus Code Challenge With C language

How To Solve HackerRank’s Plus Minus Code Challenge With C language

Problem

Let’s break down the challenge into requirements:

• Link to challenge: HackerRank’s Plus Minus Code Challenge
• n(I) = length of array arr, I = 0,1 … 100
• arr = array with values between -100 ≤ arr ≤ 100
• The goal is to console.log 3 values based on the number -ve, +ve, and zeros present in the array.
• Count the number of (positive, negative, zeros) and divide it by the total length of the array to get the ratio.

Examples

// [2, 5, 7, 0,-3,-6,0]

// positive number count = 3
printf(3/7); // 0.42

// Negative number count = 2
print(2/7) // 0.28

// zero number count = 2
print(2/7)

Goal

Write a function or functions that print the ratios for positive, negative, and zero numbers in that order

Covering Our Bases

Making sure we meet the requirements:

#include <stdio.h>


int function (arr)
{
    double positives,negatives = 0;
    int zeros = 0;

    int lengthSize = sizeof(arr) / sizeof(arr[0]);
    
     if (lengthSize > 0 && lengthSize <= 100) {
          // continue 
     } 

     // output
     print((positives / lengthSize) || 0);
     print((negatives / lengthSize) || 0);
     print((zeros / lengthSize) || 0);      
}

Understanding The Problem

This is quite a simple problem. All we need to do is count the number of positives, negatives, and zeros. In order to traverse the array, we’ll use the map function.

void plusMinus(int arr_count, int* arr) {
    int positive = 0, negative = 0, zero = 0;

    for (int i = 0; i < arr_count; i++) {
        if (arr[i] > 0) {
            positive++;
        } else if (arr[i] < 0) {
            negative++;
        } else {
            zero++;
        }
    }

    double ratioPositive = (double) positive / arr_count;
    double ratioNegative = (double) negative / arr_count;
    double ratioZero = (double) zero / arr_count;

    printf("%.6f\n", ratioPositive);
    printf("%.6f\n", ratioNegative);
    printf("%.6f\n", ratioZero);
            
}

Test Cases

Now let’s validate the code:

Input (stdin)
6
-4 3 -9 0 4 1
Your Output (stdout)
0.500000
0.333333
0.166667
Expected Output
0.500000
0.333333

Feedback

If you have any tips on how this can be better optimized or talk about coding, I would love to talk.

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