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.

  1. Case 1: Asterisk (*) character pattern follows a sequence from 0, and adding two more asterisks in each row, where row is equal to N.
  2. 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….
  3. 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++

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// 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; 

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Java

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// 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

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Output: 

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

 

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

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Improved By : offbeat