Programa para imprimir el patrón 1020304017018019020 **50607014015016 ****809012013 ******10011…

Dado un número entero N , la tarea es imprimir el siguiente patrón para el valor dado de N .  

Para N = 5, a continuación se muestra el patrón dado: 
 

Ejemplos:  

Entrada: N = 4 
Salida: 
1020304017018019020 
**50607014015016 
****809012013 
******10011

Entrada: N = 3 
Salida: 
10203010011012 
**4050809 
****607 

Enfoque: La idea para comprender la lógica detrás del patrón dado se establece a continuación:

Al mirar de cerca, vemos que al reemplazar los ceros intermedios con espacios, el patrón se puede ver más claramente. El patrón se divide además en tres patrones diferentes.

1. Caso 1: el patrón de caracteres de asterisco (*) sigue una secuencia desde 0 y agrega dos asteriscos más en cada fila, donde la fila es igual a N.
2. Caso 2: En esta parte, el patrón es muy sencillo de entender. es decir, el número de columnas y filas será igual a N y seguirá una secuencia como 1, 2, 3, 4, 5…
3. Caso 3: La secuencia de seguimiento o de abajo hacia arriba es la parte interesante donde los números se representan de abajo hacia arriba .

A continuación se muestra la implementación del enfoque anterior: 

// 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 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 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# 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

<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>

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

Complejidad de Tiempo: O(N ² )  
Espacio Auxiliar: O(1)