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Pattern Printing question asked in CGI Coding Round

  • Difficulty Level : Hard
  • Last Updated : 26 May, 2021
Geek Week

Write a program that receives a number as input and prints it in the following format as shown below. 
Examples :  

Input : n = 3
Output :
1*2*3*10*11*12
--4*5*8*9
----6*7

Input : n = 4
Output :
1*2*3*4*17*18*19*20
--5*6*7*14*15*16
----8*9*12*13
------10*11

Asked in CGI coding round

Approach: The approach is to see the problem, not as a single task but three tasks which, on combining, complete the main task. The three tasks are printing the left-half of the pattern, printing dashes(-), and printing the right-half of the pattern. Combining all three tasks, we would be able to print the pattern. 

left-half of pattern
1*2*3*
--4*5*
----6*

A function printdashes() to print the "-".

right-half of

 pattern
10*11*12
*8*9
7

Below is the implementation.

C++




// 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;
}

Java




// Java program to print the given pattern
 
class GFG {
     
// utility function to print "-" in every
// row. This will take care of printing
// "-" in the start of every row
static void printdashes(int k)
{
    int i;
    for (i = 1; i <= k; i++)
    System.out.print("-");
}
 
// function to print the pattern
static 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 = 0;
 
    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)
        System.out.print("*");
        else {
        System.out.print(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)
        System.out.print("*");
        else {
        if (k == 1)
            tracker = temp;
        System.out.print(temp);
        temp++;
        }
    }
 
    decrementor += 2;
    temp = tracker - support;
    support--;
 
    // In every row, the number of dash counts
    // is increased by 2
    dash_counter += 2;
    System.out.print("\n");
    }
}
 
// Driver code
public static void main(String arg[]) {
    int n = 3;
    pattern(n);
}
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python program to print
# the given pattern
  
# utility function to
# print "-" in every
# row. This will take
# care of printing
# "-" in the start of every row
def printdashes(k):
 
    for i in range(1,k+1):
        print("-",end="")
 
  
# function to print the pattern
def pattern(n):
      
    # variables for vertical left half
    '''
    1*2*3*
    --4*5*
    ----6*
    '''
     
    dashes = 0
    dash_counter = 0
    value = 1
  
    # variables for vertical right half
    '''
    10*11*12
    *8*9
    7
    '''
    decrementor = 0
    column_decrementor = 0
    support = n - 1
    temp = ((n * n) + 1)
    temp1 = (n * 2) - 1
    z = temp
  
    for i in range(1,n+1):
         
        printdashes(dash_counter)
          
        # This part will take
        # care of the vertical
        # left half of the pattern
        for j in range(1,(((2 * n) - dash_counter)+1)):
              
            # Printing the "*" in even positions
            if (j % 2 == 0):
                print("*",end="")
            else:
                print(value,end="")
                value=value+1
             
  
        # This part will take
        # care of the vertical
        # right half of the pattern
        for k in range(1,((temp1 - decrementor)+1)):
              
            # Printing the "*" in even positions
            if (k % 2 == 0):
                print("*",end="")
            else:
                if (k == 1):
                    tracker = temp
                print(temp,end="")
                temp=temp + 1
             
         
        decrementor =decrementor + 2
        temp = tracker - support
        support=support - 1
  
        # In every row, the number of dash counts
        # is increased by 2
        dash_counter =dash_counter + 2
        print("")
     
# driver program
n = 3
pattern(n)
 
# This code is contributed
# by Anant Agarwal.

C#




// C# program to print the given pattern
using System;
class GFG {
     
// utility function to print "-" in every
// row. This will take care of printing
// "-" in the start of every row
static void printdashes(int k)
{
    int i;
    for (i = 1; i <= k; i++)
    Console.Write("-");
}
 
// function to print the pattern
static void pattern(int n) {
 
    // variables for vertical left half
    /*
    1*2*3*
    --4*5*
    ----6*
    */
    int i, j, dash_counter = 0;
    int value = 1;
 
    // variables for vertical right half
    /*
    10*11*12
    *8*9
    7
    */
    int k, decrementor = 0;
    int support = n - 1;
    int temp = ((n * n) + 1);
    int temp1 = (n * 2) - 1;
     
 
    int tracker = 0;
 
    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)
        Console.Write("*");
        else {
        Console.Write(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)
        Console.Write("*");
        else {
        if (k == 1)
            tracker = temp;
        Console.Write(temp);
        temp++;
        }
    }
 
    decrementor += 2;
    temp = tracker - support;
    support--;
 
    // In every row, the number of dash counts
    // is increased by 2
    dash_counter += 2;
  Console.WriteLine();
    }
}
 
// Driver code
public static void Main() {
    int n = 3;
    pattern(n);
}
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to print the given pattern
 
// function to print the pattern
function pattern($n)
{
     
    // variables for vertical left half
    $dashes = 0;
    $dash_counter = 0;
    $value = 1;
 
    // variables for vertical right half
    $decrementor = 0;
    $column_decrementor = 0;
    $support = $n - 1;
    $temp = (($n * $n) + 1);
    $temp1 = ($n * 2) - 1;
    $z = $temp;
    for ($i = 1; $i <= $n; $i++)
    {
        // loop for printing dash
        for ($dd = 1; $dd <= $dash_counter;
                                     $dd++)
            printf("-");
         
        // 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 ($j % 2 == 0)
                printf("*");
            else {
                printf($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 = $temp;
                printf($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 code
$n = 3;
pattern($n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// javascript program to prvar the given pattern  
// utility function to prvar "-" in every
// row. This will take care of printing
// "-" in the start of every row
 
function printdashes(k)
{
    var i;
    for (i = 1; i <= k; i++)
        document.write("-");
}
 
// function to prvar the pattern
function pattern(n) {
 
    // variables for vertical left half
    /*
    1*2*3*
    --4*5*
    ----6*
    */
    var row, column, dashes = 0;
    var i, j, dash_counter = 0;
    var value = 1;
 
    // variables for vertical right half
    /*
    10*11*12
    *8*9
    7
    */
    var k, l, decrementor = 0;
    var column_decrementor = 0;
    var support = n - 1;
    var temp = ((n * n) + 1);
    var temp1 = (n * 2) - 1;
    var z = temp;
 
    var tracker = 0;
 
    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)
                document.write("*");
                else {
                document.write(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)
                document.write("*");
            else {
                if (k == 1)
                    tracker = temp;
                document.write(temp);
                temp++;
            }
        }
 
        decrementor += 2;
        temp = tracker - support;
        support--;
 
        // In every row, the number of dash counts
        // is increased by 2
        dash_counter += 2;
        document.write("<br>");
    }
}
 
// Driver code
var n = 3;
pattern(n);
 
 
// This code is contributed by 29AjayKumar
</script>

Output : 



1*2*3*10*11*12
--4*5*8*9
----6*7

Another approach :  

C++




// 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

C




// C program to print the given pattern
#include<stdio.h>
 
// 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)
            {
                printf("-");
            }
            else
            {
                if(col == 1)
                {
                    printf("%d", x);
                }
                else if(j <= col/2 && j % 2 == 1)
                {
                    printf("%d", x);
                    x++;
                }
                else if(j > col/2 && j % 2 == 1)
                {
                    printf("%d", t);;
                    t++;
                }
                else
                {
                    printf("*");
                }
            }                
        }
        z = (z - row) + i;
        printf("\n");
    }
}
 
// Driver code
int main()
{
    int row = 3;
    printPattern(row);
    return 0;
}
 
// This code is contributed by ankurmishra1794

Java




// Java program to print the given pattern
class GFG
{
 
// function to print the pattern
static 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)
            {
                System.out.print("-");
            }
            else
            {
                if(col == 1)
                {
                    System.out.print(x);
                }
                else if(j <= col/2 && j % 2 == 1)
                {
                    System.out.print(x);
                    x++;
                }
                else if(j > col/2 && j % 2 == 1)
                {
                    System.out.print(t);
                    t++;
                }
                else
                {
                    System.out.print("*");
                }
            }                
        }
        z = (z - row) + i;
        System.out.print("\n");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int row = 3;
    printPattern(row);
}
}
 
/* This code is contributed by PrinciRaj1992 */

Python3




# Python3 program to print the given pattern
 
# Function to print the pattern
def printPattern(row):
     
    x = 1
    z = (row * row) + 1
     
    if row == 1:
        col = 1
    else:
        col = (row * 4) - 1
     
    for i in range(1, row + 1):
        t = z
         
        for j in range(1, col - ((i - 1) * 2) + 1):
            if ((i * 2) - 2 >= j):
                print("", end = "-")          
            else:
                if (col == 1):
                    print(x, end = "")
                elif (j <= col / 2 and j % 2 == 1):
                    print(x, end = "")
                    x += 1
                elif (j > col / 2 and j % 2 == 1):
                    print(t, end = "")
                    t += 1
                else:
                    print("*", end = "")
                     
        z = (z - row) + i
        print()
 
# Driver code
row = 3
 
printPattern(row)
 
# This code is contributed by shivani

C#




// C# program to print the given pattern
using System;
 
class GFG
{
 
// function to print the pattern
static 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)
            {
                Console.Write("-");
            }
            else
            {
                if(col == 1)
                {
                    Console.Write(x);
                }
                else if(j <= col/2 && j % 2 == 1)
                {
                    Console.Write(x);
                    x++;
                }
                else if(j > col/2 && j % 2 == 1)
                {
                    Console.Write(t);
                    t++;
                }
                else
                {
                    Console.Write("*");
                }
            }                
        }
        z = (z - row) + i;
        Console.Write("\n");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int row = 3;
    printPattern(row);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// javascript program to prvar the given pattern
 
 
// function to prvar the pattern
function printPattern(row)
{
    var x = 1;
    var z = (row * row) + 1;
    var col = row == 1 ? 1 : (row * 4) - 1;
                     
    for(var i = 1; i <= row; i++)
    {
        var t = z;
        for(var j = 1; j <= col -((i - 1) * 2); j++)
        {
            if ((i * 2) - 2 >= j)
            {
                document.write("-");
            }
            else
            {
                if(col == 1)
                {
                    document.write(x);
                }
                else if(j <= col/2 && j % 2 == 1)
                {
                    document.write(x);
                    x++;
                }
                else if(j > col/2 && j % 2 == 1)
                {
                    document.write(t);
                    t++;
                }
                else
                {
                    document.write("*");
                }
            }                
        }
        z = (z - row) + i;
        document.write("<br>");
    }
}
 
// Driver code
var row = 3;
printPattern(row);
 
// This code is contributed by 29AjayKumar
</script>

Output :  

1*2*3*10*11*12
--4*5*8*9
----6*7

Another approach :  

Python3




def pattern(n):
    size = n*(n+1)
     
    # prev1 will be used to keep track of last number
    # printed in left half of pattern
    prev1 = 0
     
    # prev2 will be used to keep track of last number
    # printed in right half of pattern
    prev2 = size
    for i in range(n):
         
        # print the '-'
        print('-'*(2*i), end = '')
         
        # l1 to store numbers of left half to be printed
        l1 = [j for j in range(prev1+1, prev1+n-i+1)]
         
        # l2 to store numbers of right half to be printed
        l2 = [j for j in range(prev2-(n-i)+1,prev2+1)]
         
        # combine l1 and l2 and print the list separated by *
        print(*l1+l2, sep = '*')
         
        # decrease prev2 and increase prev1
        prev2 -= (n-i)
        prev1 += (n-i)
         
     
# driver program
n = 3
pattern(n)
   
# This code is contributed
# by Akash Jain (ultrainstinct).

Output :  

1*2*3*10*11*12
--4*5*8*9
----6*7

Another approach : 

Python3




n = 4
temp_number = (n*n)+n
counter = 1
for i in range(n):
    temp_list = []
    for j in range(n-i):
        temp_list.append(counter)
        temp_list.append(temp_number-counter+1)
        counter += 1
    temp_list.sort()
    temp_list = [str(each) for each in temp_list]
    [print("--", end="") for k in range(i)]
    print("*".join(temp_list))

Output:

1*2*3*4*17*18*19*20
--5*6*7*14*15*16
----8*9*12*13
------10*11

This article is contributed by MAZHAR IMAM KHAN. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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