Program to print the pattern 1020304017018019020 **50607014015016 ****809012013 ******10011...

Given an integer N, the task is to print the below pattern for the given value of N.

For N = 5 Below is the given pattern:

Examples:

Input: N = 4
Output:
1020304017018019020
**50607014015016
****809012013
******10011

Input: N = 3
Output:
10203010011012
**4050809
****607

Approach: The idea is to understand the logic behind the given pattern is stated below:

\By looking closely we see that by replacing in-between zeroes with spaces the pattern can be seen more clearly. The pattern is further divided into three different patterns.

Case 1: Asterisk (*) character pattern follows a sequence from 0 and adding two more asterisks in each row, where the row is equal to N.
Case 2: In this part, the pattern is very simple to understand i.e the number of columns and rows will be equal to N and follows a sequence like 1, 2, 3, 4, 5…
Case 3: Follow-up or bottom-up sequence is its interesting part where the numbers are represented from bottom to top.

Below is the implementation of the above approach:

C++

// C++ implementation to print 
// the given pattern 
#include <bits/stdc++.h> 
using namespace std; 
 
// Function to find the sum of 
// N integers from 1 to N 
int sum(int n) 
{ 
    return n * (n - 1) / 2; 
} 
 
// Function to print the given 
// pattern 
void BSpattern(int N) 
{ 
    int Val = 0, Pthree = 0, 
        cnt = 0, initial; 
 
    string s = "**"; 
 
    // Iterate over [0, N - 1] 
    for (int i = 0; i < N; i++) { 
        cnt = 0; 
 
        // Sub-Pattern - 1 
        if (i > 0) { 
            cout << s; 
            s += "**"; 
        } 
 
        // Sub-Pattern - 2 
        for (int j = i; j < N; j++) { 
 
            // Count the number of element 
            // in rows and sub-pattern 2 and 3 
            // will have same rows 
            if (i > 0) { 
                cnt++; 
            } 
            // Increment Val to print the 
            // series 1, 2, 3, 4, 5 ... 
            cout << ++Val; 
            cout << 0; 
        } 
 
        // To get the first element of sub 
        // pattern 3 find the sum of first N-1 
        // elements first N-1 elements in row1 
        // previous of Sub-Pattern 2 
 
        // Finally, add the (N-1)th element 
        // i.e., 5 and increment it by 1 
        if (i == 0) { 
            int Sumbeforelast = sum(Val) * 2; 
            Pthree = Val + Sumbeforelast + 1; 
            initial = Pthree; 
        } 
 
        // Initial is used to give the initial 
        // value of the row in Sub-Pattern 3 
        initial = initial - cnt; 
 
        Pthree = initial; 
 
        // Sub-Pattern 3 
        for (int k = i; k < N; k++) { 
 
            cout << Pthree++; 
 
            // Skip printing zero at the last 
            if (k != N - 1) { 
                cout << 0; 
            } 
        } 
 
        cout << "\n"; 
    } 
} 
 
// Driver Code 
int main() 
{ 
    // Given N 
    int N = 5; 
 
    // Function Call 
    BSpattern(N); 
    return 0; 
} 

Java

// Java implementation to print 
// the given pattern 
import java.util.*;
 
class GFG{
 
// Function to find the sum of
// N integers from 1 to N
static int sum(int n)
{
    return n * (n - 1) / 2;
}
 
// Function to print the given
// pattern
static void BSpattern(int N)
{
    int Val = 0, Pthree = 0,
        cnt = 0, initial = -1;
 
    String s = "**";
 
    // Iterate over [0, N - 1]
    for(int i = 0; i < N; i++) 
    {
        cnt = 0; 
         
        // Sub-Pattern - 1 
        if (i > 0)
        { 
            System.out.print(s);
            s += "**";
        }
 
        // Sub-Pattern - 2
        for(int j = i; j < N; j++)
        {
             
            // Count the number of element 
            // in rows and sub-pattern 2 
            // and 3 will have same rows 
            if (i > 0) 
            { 
                cnt++;
            }
             
            // Increment Val to print the
            // series 1, 2, 3, 4, 5 ...
            System.out.print(++Val);
            System.out.print("0");
        }
 
        // To get the first element of sub
        // pattern 3 find the sum of first N-1
        // elements first N-1 elements in row1
        // previous of Sub-Pattern 2
 
        // Finally, add the (N-1)th element
        // i.e., 5 and increment it by 1
        if (i == 0) 
        {
            int Sumbeforelast = sum(Val) * 2;
            Pthree = Val + Sumbeforelast + 1;
            initial = Pthree;
        }
 
        // Initial is used to give the initial
        // value of the row in Sub-Pattern 3
        initial = initial - cnt;
 
        Pthree = initial;
 
        // Sub-Pattern 3
        for(int k = i; k < N; k++)
        {
            System.out.print(Pthree++);
 
            // Skip printing zero at the last
            if (k != N - 1) 
            {
                System.out.print("0");
            }
        }
        System.out.println();
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given N
    int N = 5;
     
    // Function call
    BSpattern(N);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation to print 
# the given pattern 
 
# Function to find the sum of 
# N integers from 1 to N 
def sum(n): 
     
    return n * (n - 1) // 2
  
# Function to prthe given 
# pattern 
def BSpattern(N):
     
    Val = 0
    Pthree = 0, 
    cnt = 0 
    initial = -1
    s = "**"
  
    # Iterate over [0, N - 1] 
    for i in range(N): 
        cnt = 0
  
        # Sub-Pattern - 1 
        if (i > 0): 
            print(s, end = "")
            s += "**"
          
        # Sub-Pattern - 2 
        for j in range(i, N): 
             
            # Count the number of element 
            # in rows and sub-pattern 2 and 3 
            # will have same rows 
            if (i > 0): 
                cnt += 1
              
            # Increment Val to prthe 
            # series 1, 2, 3, 4, 5 ...
            Val += 1
            print(Val, end = "")
            print(0, end = "") 
          
        # To get the first element of sub 
        # pattern 3 find the sum of first N-1 
        # elements first N-1 elements in row1 
        # previous of Sub-Pattern 2 
  
        # Finally, add the (N-1)th element 
        # i.e., 5 and increment it by 1 
        if (i == 0): 
            Sumbeforelast = sum(Val) * 2
            Pthree = Val + Sumbeforelast + 1
            initial = Pthree 
          
        # Initial is used to give the initial 
        # value of the row in Sub-Pattern 3 
        initial = initial - cnt 
  
        Pthree = initial 
  
        # Sub-Pattern 3 
        for k in range(i, N): 
            print(Pthree, end = "")
            Pthree += 1
  
            # Skip printing zero at the last 
            if (k != N - 1): 
                print(0, end = "")
                 
        print() 
         
# Driver Code 
 
# Given N 
N = 5
  
# Function call 
BSpattern(N)
 
# This code is contributed by sanjoy_62

C#

// C# implementation to print 
// the given pattern 
using System;
class GFG{
  
// Function to find the sum of
// N integers from 1 to N
static int sum(int n)
{
    return n * (n - 1) / 2;
}
  
// Function to print the given
// pattern
static void BSpattern(int N)
{
    int Val = 0, Pthree = 0,
        cnt = 0, initial = -1;
  
    String s = "**";
  
    // Iterate over [0, N - 1]
    for(int i = 0; i < N; i++) 
    {
        cnt = 0; 
          
        // Sub-Pattern - 1 
        if (i > 0)
        { 
            Console.Write(s);
            s += "**";
        }
  
        // Sub-Pattern - 2
        for(int j = i; j < N; j++)
        {
              
            // Count the number of element 
            // in rows and sub-pattern 2 
            // and 3 will have same rows 
            if (i > 0) 
            { 
                cnt++;
            }
              
            // Increment Val to print the
            // series 1, 2, 3, 4, 5 ...
            Console.Write(++Val);
            Console.Write("0");
        }
  
        // To get the first element of sub
        // pattern 3 find the sum of first N-1
        // elements first N-1 elements in row1
        // previous of Sub-Pattern 2
  
        // Finally, add the (N-1)th element
        // i.e., 5 and increment it by 1
        if (i == 0) 
        {
            int Sumbeforelast = sum(Val) * 2;
            Pthree = Val + Sumbeforelast + 1;
            initial = Pthree;
        }
  
        // Initial is used to give the initial
        // value of the row in Sub-Pattern 3
        initial = initial - cnt;
  
        Pthree = initial;
  
        // Sub-Pattern 3
        for(int k = i; k < N; k++)
        {
            Console.Write(Pthree++);
  
            // Skip printing zero at the last
            if (k != N - 1) 
            {
                Console.Write("0");
            }
        }
        Console.WriteLine();
    }
}
  
// Driver code
public static void Main(String[] args)
{
      
    // Given N
    int N = 5;
      
    // Function call
    BSpattern(N);
}
}
  
// This code is contributed by shikhasingrajput

Output:

102030405026027028029030
**6070809022023024025
****10011012019020021
******13014017018
********15016

Time Complexity: O(N²)
Auxiliary Space: O(1)

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My Personal Notes

Program to print the pattern 1020304017018019020 50607014015016 809012013 ****10011…

C++

Java

Python3

C#

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