Pattern Printing question asked in CGI Coding Round

// C program to print the given pattern
#include <stdio.h>

// utility function to print "-" in every
// row. This will take care of printing 
// "-" in the start of every row
void printdashes(int k)
{
    int i;
    for (i = 1; i <= k; i++)
        printf("-");
}

// function to print the pattern
void pattern(int n){
    
    // variables for vertical left half
    /*
    1*2*3*
    --4*5*
    ----6*
    */
    int row, column, dashes = 0;
    int i, j, dash_counter = 0;
    int value = 1;

    // variables for vertical right half
    /*
    10*11*12
    *8*9
    7
    */
    int k, l, decrementor = 0; 
    int column_decrementor = 0;
    int support = n - 1;
    int temp = ((n * n) + 1);
    int temp1 = (n * 2) - 1;
    int z = temp;

    int tracker;

    
    for (i = 1; i <= n; i++) {
        printdashes(dash_counter);
        
        // This part will take care of the vertical
    // left half of the pattern
        for (j = 1; j <= (2 * n) - dash_counter; j++) {
            
            // Printing the "*" in even positions
            if (j % 2 == 0)
                printf("*");
            else {
                printf("%d", value);
                value++;
            }
        }

        // This part will take care of the vertical
        // right half of the pattern
        for (k = 1; k <= (temp1 - decrementor); k++) {
            
            // Printing the "*" in even positions
            if (k % 2 == 0)
                printf("*");
            else {
                if (k == 1)
                    tracker = temp;
                printf("%d", temp);
                temp++;
            }
        }
        decrementor += 2;
        temp = tracker - support;
        support--;

        // In every row, the number of dash counts
        // is increased by 2
        dash_counter += 2;
        printf("\n");
    }
}

// driver program
int main()
{   
    int n = 3;
    pattern(n);
    return 0;
}

// C++ program to print the given pattern 
#include <iostream>
using namespace std;

// function to print the pattern 
void printPattern(int row) 
{ 
    int x = 1;
    int z = (row * row) + 1;
    int col = row == 1 ? 1 : (row * 4) - 1;
    
    for(int i = 1; i <= row; i++)
    {
        int t = z;
        for(int j = 1; j <= col - ((i - 1) * 2); j++)
        {
            if ((i * 2) - 2 >= j)
            {
                cout << "-";
            }
            else
            {
                if(col == 1)
                {
                    cout << x;
                }
                else if(j <= col/2 && j % 2 == 1)
                {
                    cout << x;
                    x++;
                }
                else if(j > col/2 && j % 2 == 1)
                {
                    cout << t;
                    t++;
                }
                else
                {
                    cout << "*";
                } 
            }                 
        }
        z = (z - row) + i;
        cout << "\n";
    }
} 

// Driver code 
int main() 
{ 
    int row = 3; 
    printPattern(row); 
    return 0;
} 

// This code is contributed by shubhamsingh10

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

void pattern(int n) {
    int size = n * (n + 1);

    // prev1 will be used to keep track of last number
    // printed in left half of pattern
    int prev1 = 0;

    // prev2 will be used to keep track of last number
    // printed in right half of pattern
    int prev2 = size;
    for (int i = 0; i < n; i++) {

        // print the '-'
        for (int j = 0; j < 2 * i; j++) cout << "-";

        // l1 to store numbers of left half to be printed
        vector<int> l1;
        for (int j = prev1 + 1; j <= prev1 + n - i; j++) l1.push_back(j);

        // l2 to store numbers of right half to be printed
        vector<int> l2;
        for (int j = prev2 - (n - i) + 1; j <= prev2; j++) l2.push_back(j);

        // combine l1 and l2 and print the list separated by *
        for (int j = 0; j < l1.size(); j++) cout << l1[j] << "*";
        for (int j = 0; j < l2.size(); j++) cout << l2[j] << "*";
        cout << endl;

        // decrease prev2 and increase prev1
        prev2 -= (n - i);
        prev1 += (n - i);
    }
}

// driver program
int main() {
    int n = 3;
    pattern(n);
    return 0;
}

// This code is contributed by poojaagarwal2.

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main()
{
    int n = 4;
    int temp_number = (n * n) + n;
    int counter = 1;

    // loop through each row of the pattern
    for (int i = 0; i < n; i++) {
        vector<int> temp_list;
        // loop through each column in the current row
        for (int j = 0; j < n - i; j++) {
            // generate two numbers and add them to the
            // temp_list
            temp_list.push_back(counter);
            temp_list.push_back(temp_number - counter + 1);
            counter += 1;
        }
        // sort the numbers in the current row in ascending
        // order
        sort(temp_list.begin(), temp_list.end());
        // print the appropriate number of dashes before the
        // row
        for (int k = 0; k < i; k++) {
            cout << "--";
        }
        // print the numbers in the current row, separated
        // by asterisks
        for (int num : temp_list) {
            cout << num;
            if (num != temp_list.back()) {
                cout << "*";
            }
        }
        cout << endl;
    }
    return 0;
}
// This code is contributed by divyansh2212