Transforming Arrays with Repeated Steps and Modulo Operations in C++

Introduction:

Arrays are fundamental data structures in computer science, offering a convenient way to store and manipulate collections of elements. In certain scenarios, we encounter problems where we need to transform an array through repeated steps, incorporating specific rules. This article explores such a scenario where, given an array of integers, we perform a series of steps involving modulo operations and element additions.

Problem Statement:

Consider an array A of size N. We are tasked with repeating a specified set of steps (T times) on each element of the array. The steps involve incrementing each element by one and applying a modulo operation with the value 81. Additionally, if an element in the array becomes equal to 80 at any point, a new element with the value 0 is appended to the end of the array. The objective is to determine the size of the array after performing these transformations.

Why To Transform Arrays?

Arrays are transformed for critical reasons in programming and data analysis. This process is essential for problem-solving, data processing, algorithmic operations, normalization, preprocessing, pattern recognition, efficiency improvements, mathematical operations, and adapting to specific constraints. These transformations help simplify problems, standardize data, optimize algorithms, and reveal patterns, ultimately contributing to effective data analysis and efficient algorithmic solutions.

Many algorithmic problems require transforming arrays to meet specific conditions or constraints. The transformation process can simplify the problem-solving approach.

In data science and analysis, arrays may need to be transformed to facilitate processing, aggregation, or extraction of meaningful information.

Examples:

Example 1:

N = 4

A = [65, 2, 80, 4]

T = 3

Output: 5

Explanation:

Step 1: Initial Array

[65, 2, 80, 4]

Step 2: Iteration 1

• Increment each element by 1.
• Apply modulo 81 to each element.
• If an element becomes 80, add a new element 0 to the end.
• After the first iteration:

[66, 3, 0, 5, 1]

• The size increased to 5 due to the new element appended when the element 80 was encountered.

Step 3: Iteration 2

• Increment each element by 1.
• Apply modulo 81 to each element.
• If an element becomes 80, add a new element 0 to the end.
• After the second iteration:

[67, 4, 1, 6, 2]

• No new elements were added in this iteration, so the size remains 5.

Step 4: Iteration 3

• Increment each element by 1.
• Apply modulo 81 to each element.
• If an element becomes 80, add a new element 0 to the end.
• After the third iteration:

[68, 5, 2, 7, 3]

• Again, no new elements were added, and the size remains 5.

Example 2:

N = 4

A = {80, 80, 79, 79}

T = 2

Step 1: Initial Array

[80, 80, 79, 79]

Step 2: Iteration 1

• Increment each element by 1.
• Apply modulo 81 to each element.
• If an element becomes 80, add a new element 0 to the end.
• After the first iteration:

[0, 0, 80, 80, 0, 0]

• Here, two new elements are appended to the array because two elements became 80 during the first iteration.

Step 3: Iteration 2

• Increment each element by 1.
• Apply modulo 81 to each element.
• If an element becomes 80, add a new element 0 to the end.
• After the second iteration:

[1, 1, 0, 0, 1, 1, 0, 0]

• During the second iteration, the array undergoes further transformations. The two existing 80s are replaced by 0s, and new elements are added as some elements reach 80 again. The final array size is now 8.

After two iterations of the described transformations, the final array is [1, 1, 0, 0, 1, 1, 0, 0], and the size of the array is 8. The process involves incrementing each element by one, applying a modulo operation with the value 81, and appending new elements with the value 0 whenever an element reaches 80. This example illustrates how the array's size can grow dynamically based on the specified rules and the number of iterations.

Code:

Output:

Final Array Size: 8

Explanation:

1. calculateFinalArraySize Function:

• This function is designed to calculate the size of the array after a specified number of steps (totalSteps).
• The array frequencyCount is used to keep track of the frequency of each element in the input array.
• The loop at the beginning counts the initial frequency of each element in the input array (elements).
• The second loop iterates through the specified number of steps (totalSteps).
• Inside this loop, it updates the frequencies based on the rules described in the problem statement.
• It uses a temporary variable to swap frequencies, ensuring accurate updates.
• For element 0, it updates its frequency with the frequency of the previous element (80 in this case).
• For element 1, it increments its frequency by the frequency of element 0.
• It applies the modulo operation to avoid integer overflow.
• After the second loop, it calculates the sum of all elements in the frequencyCount array, considering the modulo operation.

2. main Function:

• The main function sets up the input parameters:
• arraySize: It is the size of the input array.
• inputArray: he input array containing integers.
• totalSteps: The number of steps to be performed.
• After that, it calls the calculateFinalArraySize function with these parameters and prints the result.

Note: The const int modulo = 1e9 + 7; is a constant to define the modulo operation to avoid integer overflow.

• The array frequencyCount is used to efficiently keep track of the frequency of each element.
• The calculateFinalArraySize function encapsulates the logic of updating frequencies and calculating the final array size after a certain number of steps.
• In the provided example, the main function calls calculateFinalArraySize with the input array [80, 80, 79, 79], and after 2 steps, it outputs the final array size, which is 8.

Time and Space Complexities:

Time Complexity:

• Frequency Counting: In the calculateFinalArraySize function, the loop for counting the initial frequency of each element runs in O(N) time, where N is the size of the input array.
• Performing T Steps: The loop for performing T steps has a constant upper limit of 81 iterations, which doesn't depend on the size of the input array. Therefore, it contributes O(1) to the time complexity.
• Total Time Complexity: The overall time complexity is dominated by the initial frequency counting, making it O(N).

Space Complexity:

• Frequency Count Array: The frequencyCount array is used to store the frequency of each element. It has a constant size of 82 (81 elements + 1 extra for the additional 0 during swapping). Therefore, it contributes O(1) to the space complexity.
• Input Vector: The elements vector, representing the input array, contributes O(N) to the space complexity, where N is the size of the input array.
• Other Variables: The remaining variables and constants are of fixed size and don't depend on the input size. Therefore, they contribute O(1) to the space complexity.

Time Complexity: O(N)

Space Complexity: O(N)

Conclusion:

In summary, the C++ implementation efficiently addresses the problem of transforming arrays with repeated steps and modulo operations. The algorithm employs a frequency count array for tracking element occurrences, ensuring a time complexity of O(N) and a space complexity of O(N), where N is the array size. The code exhibits clarity, readability, and a balanced approach to handling dynamic array transformations, making it suitable for practical applications with moderate-sized arrays.