HackerRank Plus Minus Java Solution Explained
Problem Overview
Original Problem: Plus Minus | HackerRank
Difficulty: Easy
Category: Arrays, Math
Description: Given an array of integers, compute the ratios of positive, negative, and zero elements. Output each ratio as a decimal with six places, printed on separate lines.
This problem is part of HackerRank’s One Week Preparation Kit, designed to help beginners prepare for coding interviews by practicing fundamental array and math operations in Java.
Understanding the Problem
The Plus Minus problem is a straightforward challenge that introduces beginners to array traversal, basic counting, and handling decimal precision in Java. The goal is to analyze an array of integers and determine the proportion of positive numbers (greater than 0), negative numbers (less than 0), and zeros. Each proportion is calculated by dividing the count of each category by the total number of elements, then formatting the result to six decimal places.
Input and Output
Input:
First line: An integer n (the size of the array, 1 ≤ n ≤ 100).
Second line: n space-separated integers, each in the range [-100, 100].
Output:
Three lines, each containing a decimal value (to six decimal places):
Proportion of positive numbers.
Proportion of negative numbers.
Proportion of zeros.
Initial Reasoning
To solve this, we need to:
Count the occurrences of positive, negative, and zero elements in the array.
Divide each count by the array length to get the ratios.
Format the output to six decimal places.
A single pass through the array is sufficient, making the time complexity O(n). The challenge also tests precision handling, as floating-point calculations must be formatted correctly to meet the six-decimal-place requirement.
Thought Process
Let’s break down the solution logically:
Traverse the Array: Iterate through the array once, checking each element to determine if it’s positive, negative, or zero. Use counters for each category.
Calculate Ratios: Divide the counts by the array length to compute the proportions. Since the array length is guaranteed to be at least 1, we don’t need to worry about division by zero.
Handle Precision: Use Java’s formatting tools (e.g., System.out.printf) to ensure the output is displayed with six decimal places.
Edge Cases:
If the array has only one element, the ratio for that element’s category will be 1.0, and others will be 0.0.
If a category has no elements (e.g., no positives), the ratio will be 0.0.
The constraints ensure all inputs are valid integers, so no need to handle non-integer or out-of-range inputs.
This approach is optimal because it uses O(1) extra space (just three counters) and O(n) time for a single array pass.
Pseudocode
// plusMinus method:
Initialize counters for positive, negative, and zero elements
Get the array length
For each element in the array
If the element is positive
Increment the positive counter
If the element is negative
Incremement the negative counter
Else
Incremement the zero counter
Calculate the ratios
Print each ratio formatted to 6 decimal places
Solution Code
// Note: This is one of many possible solutions
// Note: Variable names have been updated from the original template for clarity and to follow Java naming conventions (e.g., 'n' to 'arrayLength')
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
public static void plusMinus(List<Integer> numbers) {
int positiveCount = 0;
int negativeCount = 0;
int zeroCount = 0;
int arrayLength = numbers.size();
// Count positive, negative, and zero elements
for (int num : numbers) {
if (num > 0) {
positiveCount++;
} else if (num < 0) {
negativeCount++;
} else {
zeroCount++;
}
}
// Calculate ratios
double positiveRatio = (double) positiveCount / arrayLength;
double negativeRatio = (double) negativeCount / arrayLength;
double zeroRatio = (double) zeroCount / arrayLength;
// Print ratios with six decimal places
System.out.printf("%.6f%n", positiveRatio);
System.out.printf("%.6f%n", negativeRatio);
System.out.printf("%.6f%n", zeroRatio);
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
int arrayLength = Integer.parseInt(bufferedReader.readLine().trim());
List numbers = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
.map(Integer::parseInt)
.collect(toList());
Result.plusMinus(numbers);
bufferedReader.close();
}
} Step-by-Step Explanation
Let’s walk through the Java solution to ensure beginners understand each part:
Input Handling:
The main method reads the input using BufferedReader.
The first line is the array size, converted to an integer with Integer.parseInt.
The second line is a space-separated list of integers, processed using Stream.of and map to create a List<Integer>.
Counting Elements:
In the plusMinus method, we initialize three counters: positiveCount, negativeCount, and zeroCount.
We store the array size in arrayLength using numbers.size() for clarity and to avoid repeated calls.
A for-each loop iterates through the array, checking each element:
If num > 0, increment positiveCount.
If num < 0, increment negativeCount.
If num == 0, increment zeroCount.
Calculating Ratios:
Compute ratios by dividing each count by arrayLength. We cast the numerator to double (e.g., (double) positiveCount) to ensure floating-point division.
For example, if positiveCount = 3 and arrayLength = 5, then positiveRatio = 3.0 / 5 = 0.6.
Formatting Output:
Use System.out.printf("%.6f%n", ratio) to print each ratio with six decimal places.
The %.6f format specifier ensures six decimal places, and %n adds a newline.
This handles the precision requirement (absolute error up to 10^-4 is acceptable).
Closing Resources:
The BufferedReader is closed to prevent resource leaks.
Common Mistakes to Avoid
Here are pitfalls beginners often encounter:
Using Integer Division: Dividing integers without casting to double results in integer division, which truncates the decimal part. Always cast to double for accurate ratios.
Incorrect Formatting: Forgetting to use %.6f or using println instead of printf can lead to incorrect decimal precision.
Misinterpreting Constraints: Assuming the array is empty or contains non-integers. The constraints (1 ≤ n ≤ 100, -100 ≤ arr[i] ≤ 100) ensure valid inputs, so no extra validation is needed.
Real-World Application
The logic behind Plus Minus is common in data analysis and reporting. For example, a business might analyze customer feedback scores (positive, negative, neutral) to compute proportions for a dashboard. Similarly, in machine learning, you might calculate the distribution of data points across categories to understand dataset balance. This problem teaches foundational skills in array processing and precision handling, which are critical in data science and software development.
FAQs
Q: Why use System.out.printf instead of System.out.println?
A: printf allows precise control over decimal places (e.g., %.6f for six decimals), which is required by the problem. println doesn’t offer this formatting.
Q: Can I use Java streams to solve this?
A: Yes, but for beginners, a simple loop is more readable and equally efficient. Streams might look like: arr.stream().filter(x -> x > 0).count().
Q: How do I handle large arrays?
A: The constraints (n ≤ 100) ensure arrays are small, so a single loop is sufficient. For larger arrays, the same logic applies, as the solution scales linearly (O(n)).