Buggy Code Array Sum in C++

Inside the code editor we've tried to write a function that takes an array nums as argument and prints to the console the sum of all numbers in that list.

So when we called printSum({1, 2, 3}), we expected our code to print:

because 1 + 2 + 3 = 6. But it seems like we made some mistakes because when we run our code, it prints:

1
2
3

Assignment:

Your task is to fix our function such that it correctly computes and prints the desired sum.

Understanding the Problem

The core challenge of this problem is to correctly compute the sum of all elements in an array and print the result. The initial code seems to print each element of the array individually rather than summing them up.

This problem is significant as it tests basic understanding of array manipulation and summation in C++. Such operations are common in various applications, including data analysis, statistics, and algorithm design.

Potential pitfalls include misunderstanding the loop structure or incorrectly accessing array elements.

Approach

To solve this problem, we need to iterate through the array, compute the sum of its elements, and then print the sum. Let's break down the steps:

Initialize a variable to store the sum.
Iterate through each element of the array.
Add each element to the sum variable.
After the loop, print the sum.

Initially, a naive solution might involve printing each element, but this does not meet the requirement. Instead, we need to accumulate the sum and print it once.

Algorithm

Here is a step-by-step breakdown of the algorithm:

Initialize a variable sum to 0.
Use a loop to iterate through each element of the array.
In each iteration, add the current element to sum.
After the loop ends, print the value of sum.

Code Implementation


#include <iostream>
#include <vector>

void printSum(const std::vector<int>& nums) {
    // Initialize sum to 0
    int sum = 0;
    
    // Iterate through each element in the array
    for (int num : nums) {
        sum += num; // Add each element to sum
    }
    
    // Print the final sum
    std::cout << sum << std::endl;
}

int main() {
    // Test the function with an example input
    std::vector<int> nums = {1, 2, 3};
    printSum(nums); // Expected output: 6
    return 0;
}

Complexity Analysis

The time complexity of this approach 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 only use a single integer variable to store the sum.

Edge Cases

Consider the following edge cases:

An empty array: The sum should be 0.
An array with one element: The sum should be the element itself.
An array with negative numbers: The sum should correctly account for negative values.

Examples:


printSum({}) // Expected output: 0
printSum({5}) // Expected output: 5
printSum({-1, -2, -3}) // Expected output: -6

Testing

To test the solution comprehensively, consider a variety of test cases:

Simple cases with small arrays.
Edge cases as mentioned above.
Large arrays to ensure performance.

Using a testing framework like Google Test can help automate and manage these tests effectively.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Write pseudo-code to outline your approach before coding.
Test your solution with different inputs to ensure it handles all cases.
Practice similar problems to improve your problem-solving skills.

Conclusion

In this post, we discussed how to fix a buggy function to correctly compute and print the sum of an array in C++. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong programming skills.

Keep practicing and exploring further to enhance your problem-solving abilities!

Additional Resources

For further reading and practice, consider the following resources:

Array Sum: Buggy Code in C++ (Time Complexity: O(n))