Introduction:
Alphanumeric patterns dissect their structured grid to combine letters and numbers such as puzzles. These types of patterns tend to require programmers to create pattern recognition and generation algorithms within them. For C++ developers, solving these patterns does not only help them sharpen coding skills but also improve logical thinking and algorithmic understanding. This article majorly focuses on one particular alpha numerical pattern from Matrix I and how it can be implemented through using C++.
Problem Statement:
Consider the following alphanumerical pattern:
Example
1A 2B 3C 4D 5E
2F 3G 4H 5I 6J
3K 4L 5M 6N 7O
4P 5Q 6R 7S 8T
5U 6V 7W 8X 9Y
The task is to write a C++ program that generates this pattern. The pattern follows a specific rule:
- The first number in each row should increase by 1.
- Each column starts with an alphabet letter which proceeds alphabetically.
- These numbers and letters have been mixed together as shown.
Approach:
In order to generate the alphanumerical pattern, we can follow a structured approach in C++. Here are the steps we will take:
- Declare variables and constants: We need to declare variables for keeping track of how many rows or columns are allocated so we can use them later while constants are required for defining the size of the pattern and range of characters used.
- Write nested loops: To generate pattern elements, the nest loop will be written using increased loops over rows then columns.
- Calculate and print out the pattern elements: In between those brackets, numbers & letters are determined subjecting themselves under specific rules of that bidirectional trend then outputting into that targeted sequence.
Implementation in C++ (Program 1):
Let's now translate the approach into C++ code:
Example
#include <iostream>
using namespace std;
int main() {
const int rows = 5;
const int cols = 5;
char letter = 'A';
for (int i = 0; i < rows; i++) {
int number = i + 1;
for (int j = 0; j < cols; j++) {
if (j == 0) {
cout << number;
} else {
cout << letter++;
}
cout << " ";
}
letter = 'A'; // Reset letter for the next row
cout << endl;
}
return 0;
}
Output:
Output
1 A B C D
2 A B C D
3 A B C D
4 A B C D
5 A B C D
Explanation:
In this C++ program:
- Initialization:
- The first task of the program is to initialize the pattern constants: rows and cols.
- Also the letter gets initialized by 'A' , which is the starting character of the pattern.
- Outer Loop (Rows):
- In this loop, the program starts from 0 which is the index of first row and stops at rows - 1 which represents the last row.
- For each row, it calculates a number that corresponds with its position in the list. The pattern begins with 1 and adds one to every other line thus uses number = i + 1 where i refers to the current row index.
- Inner Loop (Columns):
- Within this outer loop for rows, there is an inner loop for columns. It goes through all columns for a given row starting at column zero and going until cols-1 for that last column.
- Inside this loop, there is a test if "current column" equals "first column" (index 0). In case it does print out this calculated "number" otherwise prints out the current "letter" then increases the letter by one alphabet letter.
- Printing the Pattern:
- Numbers and letters are printed as per rule of pattern once every column has been traversed in this inside loop.
- After printing all elements of the current row, the program moves to the next line (endl), indicating the end of that row.
- Resetting Letter:
- Before moving to the next row in the outer loop, the program resets the letter variable back to 'A'. This ensures that each new row starts with the letter 'A' again, following the pattern's structure.
By repeating the above steps for each row and column, the program generates the desired alphanumerical pattern as specified in Matrix I.
Time Complexity:
- Two nested loops, one for rows and another for cols each iterating from 0 to a constant value (rows and cols) is all this program demands.
- Therefore the time complexity can be expressed as O(rows * cols).
Space Complexity:
- The program uses a few variables such as rows, cols, letter, number, i and j which all require constant space regardless of the input size.
- Additionally, the program prints the pattern directly to the standard output without storing it in any data structure.
- Hence, the space complexity remains constant and it is denoted by O(1).
Program 2:
Let us take another example to illustrate how to match the alphanumerical pattern in matrix I in C++.
Example
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
// Function to check if the pat matches at a given
// starting position in the mat.
bool matchPattern(const vector<vector<int> >& mat,
const vector<string>& pat, int row,
int col)
{
const int m1 = mat.size(), n1 = mat[0].size(),
m2 = pat.size(), n2 = pat[0].length();
// Create a map to store the mapping between characters
// in the pat and their corresponding values in the
// mat.
unordered_map<char, int> charToValue;
// Iterate over the rows and columns of the pat.
for (int i = 0; i < m2; ++i) {
for (int j = 0; j < n2; ++j) {
const char c = pat[i][j];
const int v = mat[row + i][col + j];
// Check if the character is a digit.
if (isdigit(c)) {
// If the character is a digit, it must
// match the corresponding value in the
// mat.
if (c != v + '0') {
return false;
}
}
else {
// If the character is not a digit, it must
// have been mapped to a value in the mat
// previously.
if (charToValue.count(c)) {
// If the character has been mapped to a
// different value, it's a mismatch.
if (charToValue[c] != v) {
return false;
}
}
else {
// If the character has not been mapped
// before, map it to the current value.
charToValue[c] = v;
}
}
}
}
return true; // Return true if the pat matches at
// the given starting position.
}
vector<int> findPattern(vector<vector<int> >& mat,
vector<string>& pat)
{
const int m1 = mat.size(), n1 = mat[0].size(),
m2 = pat.size(), n2 = pat[0].length();
// Iterate over all possible starting positions for the
// pat in the mat.
for (int i = 0; i + m2 <= m1; ++i) {
for (int j = 0; j + n2 <= n1; ++j) {
// Check if the pat matches at the current
// starting position.
if (matchPattern(mat, pat, i, j)) {
return {
i, j
}; // Return the starting position if a
// match is found.
}
}
}
return { -1,
-1 }; // Return {-1, -1} if no match is found.
}
// Main code
int main()
{
vector<vector<int> > mat
= { { 1, 2, 2 }, { 2, 2, 3 }, { 2, 3, 3 } };
vector<string> pat = { "ab", "bb" };
;
// Call the findPattern method to find the starting
// position of the pat in the mat.
vector<int> result = findPattern(mat, pat);
// Print the result.
cout << "Starting position of the pat: (" << result[0]
<< ", " << result[1] << ")" << endl;
return 0;
}
Output:
Output
Starting position of the pat: (0, 0)
Explanation:
- Iterate over possible starting positions:
- In this example, loop through all possible starting positions for the pat in the mat.
- For each starting position:
- Call the matchPattern function to check if the pat matches at that position.
- If the pat matches, return the starting position.
- If no match is found, return {-1, -1}.
- In matchPattern function, iterate over pat rows and columns:
- Loop through each row and column of the pat.
- For each character in the pat:
- If the character is a digit: Check if the corresponding value in the mat matches the digit.
- If the character is not a digit: Check if the character has been mapped to a value in the mat previously.
- If not, map the character to the corresponding value in the mat.
- If the character has been mapped to a different value, return false.
- If all characters in the pat match, return true.
Conclusion:
To conclude we have seen how we can generate alpha-numerical pattern Matrix I using above c++ code. This exercise not only demonstrates nested loops and control structures in C++ but also programming concepts used to solve such puzzling pattern problems. Modifying this pattern or working with more complex patterns aids in improving programming skills and promotes creative thinking in problem-solving.