Inside the code editor we've tried to write a function that takes an array nums
as argument and for every number in the array:
prints "negative" if that number is less than zero
or prints the number itself, if the number is greater than or equal to zero
So when we called the function for [2, -12, 0, 4, -5, 3]
, we expected it to print:
2
negative
0
4
negative
3
but it seems like we made some mistakes because when we run our code, it produces this output:
positive
-12
0
positive
-5
positive
Assignment:
Your task is to fix our loop such that it will print the desired output.
The core challenge of this problem is to correctly identify and print whether each number in the array is negative or non-negative. The significance of this problem lies in its simplicity and the fundamental understanding of conditional statements in programming. A common pitfall is misunderstanding the conditions or incorrectly implementing the logic, leading to incorrect outputs.
To solve this problem, we need to iterate through the array and check each number. If the number is less than zero, we print "negative". Otherwise, we print the number itself. The initial naive solution might involve incorrect conditions or print statements, as seen in the provided buggy code.
The naive solution might look like this:
for (int num : nums) {
if (num > 0) {
System.out.println("positive");
} else {
System.out.println(num);
}
}
This solution is incorrect because it prints "positive" for numbers greater than zero, which is not what we want. We need to print the number itself if it is non-negative.
The optimized solution involves correctly checking if the number is less than zero and printing "negative" in that case. Otherwise, we print the number itself.
Here is the step-by-step breakdown of the optimized algorithm:
public class PrintPositiveNegative {
public static void main(String[] args) {
int[] nums = {2, -12, 0, 4, -5, 3};
printPositiveNegative(nums);
}
public static void printPositiveNegative(int[] nums) {
// Iterate through each number in the array
for (int num : nums) {
// Check if the number is less than zero
if (num < 0) {
// Print "negative" if the number is less than zero
System.out.println("negative");
} else {
// Print the number itself if it is non-negative
System.out.println(num);
}
}
}
}
The time complexity of this solution is O(n), where n is the number of elements in the array. This is because we iterate through each element exactly once. The space complexity is O(1) as we are not using any additional space that scales with the input size.
Potential edge cases include:
Example edge cases and their expected outputs:
printPositiveNegative(new int[] {}); // No output
printPositiveNegative(new int[] {-1, -2, -3}); // negative\nnegative\nnegative
printPositiveNegative(new int[] {0, 1, 2}); // 0\n1\n2
To test the solution comprehensively, we should include a variety of test cases:
Using a testing framework like JUnit can help automate and manage these tests effectively.
When approaching such problems, it is crucial to:
Practicing similar problems and studying different algorithms can significantly improve problem-solving skills.
In this blog post, we discussed how to fix a buggy Java function to correctly print "negative" for negative numbers and the number itself for non-negative numbers. We explored the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is essential for developing strong programming skills.
For further reading and practice, consider the following resources: