C++ Coding Test Course, Finding the Sum of Remainders

Problem Description

Given an integer array and an integer K, this problem requires us to find the remainder when each element of the array is divided by K and to compute the sum of these remainders. To solve this problem, we need to traverse the array, take the remainder for each element divided by K, and then accumulate these remainders for summation.

Problem Summary

• Input: Integer array arr and integer K
• Output: Sum of the remainders of the array elements when divided by K

Input Example

[
    3, 7, 2, 9, 10
]
K = 4

Output Example

Sum of remainders: 4

Problem Solving Strategy

To solve this problem, we will adopt the following strategy.

1. Store the input array and K in variables.
2. Traverse the array and find the remainder for each element when divided by K.
3. Accumulate the found remainders for summation.
4. Finally, output the sum of the remainders.

C++ Code Implementation

Below is the code implemented in C++.

#include <iostream>
#include <vector>

using namespace std;

int main() {
    vector<int> arr = {3, 7, 2, 9, 10}; // Input array
    int K = 4; // Value of K
    int remainderSum = 0; // Variable to store the sum of remainders
    
    // Traverse the array
    for (int num : arr) {
        remainderSum += num % K; // Calculate and accumulate the remainders
    }
    
    // Print the result
    cout << "Sum of remainders: " << remainderSum << endl;
    return 0;
}

Main Code Explanation

Here is a detailed explanation of each part of the code.

• #include <iostream>: Includes the iostream library for input and output functionality.
• #include <vector>: Includes the vector library to use dynamic arrays.
• using namespace std;: Uses the std namespace for cleaner code.
• vector<int> arr = {3, 7, 2, 9, 10};: Initializes the integer array to be used for calculations.
• int K = 4;: Declares the integer K for calculating remainders.
• int remainderSum = 0;: Initializes the variable to store the sum of remainders.
• for (int num : arr): Traverses all elements of the array.
• remainderSum += num % K;: Stores the remainder of each element when divided by K.
• cout << "Sum of remainders: " << remainderSum << endl;: Outputs the final result.

Time Complexity Analysis

The time complexity of this algorithm is O(N), where N represents the number of elements in the array. It shows linear performance as it traverses each element once to compute the remainder and accumulates these values.

Test Cases

Below are additional test cases designed to verify results for various inputs.

• Test Case 1:

  arr = [1, 2, 3], K = 1

  Output: 0 (The remainder of all numbers divided by 1 is 0)

• Test Case 2:

  arr = [5, 6, 7, 8], K = 3

  Output: 6 (5 % 3 = 2, 6 % 3 = 0, 7 % 3 = 1, 8 % 3 = 2, total sum = 2 + 0 + 1 + 2 = 5)

• Test Case 3:

  arr = [10, 20, 30], K = 10

  Output: 0 (The remainder of all numbers divided by 10 is 0)

Conclusion

In this lecture, we explored the process of implementing and solving the problem of finding the sum of remainders using C++. Through this problem, we can learn the fundamental flow of algorithms that perform operations while traversing an array. Such types of problems are useful for algorithm exam preparation and will be beneficial for coding tests. It is advisable to validate the code with various input cases and practice solving while maintaining performance as needed.

“Programming is the process of solving problems. Understand the problem deeply and try to find the appropriate solution.”