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Program to print numeric pattern | Set – 2

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Given a number as ‘num’, and a number of lines as ‘num_of_lines’ where ‘num’ implies the starting number from which the pattern has to be started and ‘num_of_lines’ implies the number of lines that have to be printed. Now, according to the above information, print a pattern as given below.

Examples: 

Input: num = 7, num_of_lines = 4
Output: 7
        14 15
        28 29 30 31
        56 57 58 59 60 61 62 63

Input: num = 3, num_of_lines = 3
Output: 3
        6 7
        12 13 14 15

Observations: 

  • Elements of the first column are the multiple of the previous element in that column.
  • No. of elements in each row is twice the no. of elements in the previous row.
  • Also, to generate next element in the row, add 1 to the previous element of that row.

Approach: 
So, start a loop from 0 to num_of_lines-1, to take care of the number of rows to be printed and another loop inside the first loop, from 0 till limit-1, the limit will be initialized to 1, and its value is increased exponentially. Now inside the loop, just increase the number by 1 to print the next number of that row. 
 

C++




// C++ program to print the
// given numeric pattern
#include <bits/stdc++.h>
using namespace std;
 
// Function to print th epattern
void printPattern (int num, int numOfLines )
{
    int n = num, num2, x = 1, limit = 1;
 
    // No. of rows to be printed
    for (int i = 0; i < numOfLines; i++) {
         
        // No. of elements to be printed in each row
        for (int j = 0; j < limit; j++) {
            if (j == 0)
                num2 = num;
 
            // Print the element
            cout << num2++ << " ";
        }
        num *= 2;
        limit = num / n;
        cout << endl;
    }
}
 
// Drivers code
int main()
{
 
    int num = 3;
    int numOfLines = 3;
 
    printPattern(num,  numOfLines);
     
    return 0;
}


Java




// Java program to print the
// given numeric pattern
class solution_1
{
     
// Function to print
// the pattern
static void printPattern (int num,
                          int numOfLines)
{
    int n = num, num2 = 0,
        x = 1, limit = 1;
 
    // No. of rows to
    // be printed
    for (int i = 0;
             i < numOfLines; i++)
    {
         
        // No. of elements to be
        // printed in each row
        for (int j = 0; j < limit; j++)
        {
            if (j == 0)
                num2 = num;
 
            // Print the element
            System.out.print(num2++ + " ");
        }
         
        num *= 2;
        limit = num / n;
        System.out.println();
    }
}
 
// Driver code
public static void main(String args[])
{
    int num = 3;
    int numOfLines = 3;
 
    printPattern(num, numOfLines);
}
}
 
// This code is contributed
// by Arnab Kundu


Python 3




# Python 3 program to print
# the given numeric pattern
 
# Function to print th epattern
def printPattern (num, numOfLines ):
 
    n = num
    limit = 1
 
    # No. of rows to be printed
    for i in range(0, numOfLines):
         
        # No. of elements to be
        # printed in each row
        for j in range(limit):
            if j == 0:
                num2 = num
 
            # Print the element
            print(num2, end = " ")
            num2 += 1
     
        num *= 2
        limit = num // n
        print()
         
# Driver code
if __name__ == "__main__":
 
    num = 3
    numOfLines = 3
 
    printPattern(num, numOfLines)
 
# This code is contributed
# by ChitraNayal


C#




// C# program to print the
// given numeric pattern
using System;
class GFG
{
     
// Function to print
// the pattern
static void printPattern(int num,
                         int numOfLines)
{
    int n = num, num2 = 0,
        limit = 1;
 
    // No. of rows to
    // be printed
    for (int i = 0;
             i < numOfLines; i++)
    {
         
        // No. of elements to be
        // printed in each row
        for (int j = 0; j < limit; j++)
        {
            if (j == 0)
                num2 = num;
 
            // Print the element
            Console.Write(num2++ + " ");
        }
         
        num *= 2;
        limit = num / n;
        Console.Write("\n");
    }
}
 
// Driver code
public static void Main()
{
    int num = 3;
    int numOfLines = 3;
 
    printPattern(num, numOfLines);
}
}
 
// This code is contributed by Smitha


PHP




<?php
// PHP program to print the
// given numeric pattern
 
// Function to print th epattern
function printPattern($num, $numOfLines)
{
    $n = $num;
    $limit = 1;
 
    // No. of rows to be printed
    for ($i = 0; $i < $numOfLines; $i++)
    {
         
        // No. of elements to be
        // printed in each row
        for ($j = 0; $j < $limit; $j++)
        {
            if ($j == 0)
                $num2 = $num;
 
            // Print the element
            echo $num2++ . " ";
        }
        $num *= 2;
        $limit = $num / $n;
        echo "\n";
    }
}
 
// Driver code
$num = 3;
$numOfLines = 3;
 
printPattern($num, $numOfLines);
     
// This code is contributed
// by ChitraNayal
?>


Javascript




<script>
 
  // JavaScript program to print the
 // given numeric pattern
 
  // Function to print th epattern
     function printPattern(num, numOfLines)
     {
        var n = num,
          num2,
          x = 1,
          limit = 1;
 
        // No. of rows to be printed
        for (var i = 0; i < numOfLines; i++)
        {
          // No. of elements to be
          // printed in each row
          for (var j = 0; j < limit; j++) {
            if (j == 0) num2 = num;
 
            // Print the element
            document.write(num2++ + "  ");
          }
          num *= 2;
          limit = num / n;
          document.write("<br>");
        }
      }
 
      // Drivers code
      var num = 3;
      var numOfLines = 3;
 
      printPattern(num, numOfLines);
       
 </script>


Output: 

3 
6 7 
12 13 14 15

 

Time Complexity: O(pow(2,n)), Here n is the number of lines.
Auxiliary Space: O(1), As constant extra space is used.



Last Updated : 23 Dec, 2022
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