Printing patterns

You are required to form a matrix of size $r\times c$ where $r$ is the number of rows and $c$ is the number of columns. You are required to form the waves of numbers around the provided center, $(C_i,C_j)$ (0-based indexing).

Example:

• Size of the matrix: $r=9$ and $c=9$
• Center coordinates: $C_i=4$ and $C_j=4$ (0 based indexing)
• Pattern:

            4 4 4 4 4 4 4 4 4 
            4 3 3 3 3 3 3 3 4 
            4 3 2 2 2 2 2 3 4 
            4 3 2 1 1 1 2 3 4 
            4 3 2 1 0 1 2 3 4 
            4 3 2 1 1 1 2 3 4 
            4 3 2 2 2 2 2 3 4 
            4 3 3 3 3 3 3 3 4 
            4 4 4 4 4 4 4 4 4

You are given the values of $r,c,C_i,C_j$ (the values of $C_i$ and $C_j$ is 0-based indexed). Your task is to print the provided pattern.

Example:

Input:  r=10, c=10, x=5, y=5
Output: 

5 5 5 5 5 5 5 5 5 5 
5 4 4 4 4 4 4 4 4 4 
5 4 3 3 3 3 3 3 3 4 
5 4 3 2 2 2 2 2 3 4 
5 4 3 2 1 1 1 2 3 4 
5 4 3 2 1 0 1 2 3 4 
5 4 3 2 1 1 1 2 3 4 
5 4 3 2 2 2 2 2 3 4 
5 4 3 3 3 3 3 3 3 4 
5 4 4 4 4 4 4 4 4 4 

Approach

Java

public class PrintingPatterns {
    public static void main(String args[]) {
        int r = 10;
        int c = 10;
        int x = 5;
        int y = 5;
        StringBuilder str = new StringBuilder();
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                str.append(Math.max(Math.abs(x - i), Math.abs(y - j))).append(" ");
            }
            str.append("\n");
        }
        System.out.println(str);
    }
}

ques 6

Given an integer n. Print a pyramid made up of symbol '*' of height n.

Example:

Input:  n = 3

Output:

  *
 ***
*****

Approach:

C++

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n = 3;

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n - i; j++)
            cout << " ";
        for (int k = 0; k < 2 * i - 1; k++)
            cout << "*";
        cout << "\n";
    }
    cout << "\n";

    return 0;
}

The Pattern

Joey is always fond of playing with patterns. So on his birthday, his uncle gave him a mat.  Mat is a rectangular grid

$(NM)$. i.e. there are $N$ rows and $M$ columns are drawn on the mat.  Each cell of this grid is filled with either dot$(.)$ or star$(\ast )$. Now Joey starts to fold this mat.  Firstly he folds the mat along the rows (top to down) until it transforms into a $(1\ast M)$ grid (Neglect width of mat on each fold). After that, he starts folding this along the columns (left to right) and finally transforms into a single cell. In the end Joey wants to know what will be on the top of that cell $($dot $(.)$or star$(\ast ))$.

When star$(\ast )$ come over dot$(.)$ it converted to dot$(.)$
When star$(\ast )$ come over star$(\ast )$, it remains the star$(\ast )$
When dot$(.)$ come over dot$(.)$, it remains dot$(.)$
When dot$(.)$ come over star$(\ast )$, it converted to star$(\ast )$

Example:

Input:  n = 5, m = 5 , pattern={{"*.***"},{".**.."},{".*.*."},{"*...*"},{"..*.*"}}
Output: *

Approach

C++

#include <bits/stdc++.h>
using namespace std;

void thePattern(vector<string> pattern, int n, int m)
{
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (pattern[i][j] == '.' && 
                  pattern[i + 1][j] == '*')
                pattern[i + 1][j] = '*';
            else if (pattern[i][j] == '*' &&
                      pattern[i + 1][j] == '.')
                pattern[i + 1][j] = '.';
        }
    }

    for (int j = 0; j < m - 1; j++)
    {
        if (pattern[n - 1][j] == '.' && 
                  pattern[n - 1][j + 1] == '*')
            pattern[n - 1][j + 1] = '*';
        else if (pattern[n - 1][j] == '*' &&
        pattern[n - 1][j + 1] == '.')
            pattern[n - 1][j + 1] = '.';
    }
    cout << pattern[n - 1][m - 1] << "\n";
}
int main()
{
    int n = 5, m = 5;

    vector<string> pattern = {{"*.***"},
                              {".**.."},
                              {".*.*."},
                              {"*...*"},
                              {"..*.*"}};

    thePattern(pattern, n, m);

    return 0;
}