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

Given an integer N, the task is to print the pattern below 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 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 adds two more asterisks in each row, where the 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 the 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 print the 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 print the
            # 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

Javascript

<script>
//Javascript implementation to print
// the given pattern

// Function to find the sum of
// N integers from 1 to N
function sum( n)
{
    return n * parseInt((n - 1) / 2);
}

// Function to print the given
// pattern
function BSpattern( N)
{
    var Val = 0, Pthree = 0,
        cnt = 0, initial;

    var s = "**";

    // Iterate over [0, N - 1]
    for (var i = 0; i < N; i++) {
        cnt = 0;

        // Sub-Pattern - 1
        if (i > 0) {
            document.write( s);
            s += "**";
        }

        // Sub-Pattern - 2
        for (var 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 ...
            document.write( ++Val);
            document.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) {
            var 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 (var k = i; k < N; k++) {

            document.write(Pthree++);

            // Skip printing zero at the last
            if (k != N - 1) {
                document.write( 0);
            }
        }

        document.write( "<br>");
    }
}

// Given N
var N = 5;
// Function Call
BSpattern(N);

// This code is contributed by SoumikMondal
</script>

Output:

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

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