Mountain Sequence Pattern

Given a number N, the task is to generate the pyramid sequence pattern which contains N pyramids one after the other as shown in the examples below.
Examples:

Input: N = 3
Output:
  *    *    *
 ***  ***  ***
***************

Input: N = 4
Output: 
  *    *    *    *
 ***  ***  ***  ***
********************

Iterative Approach: The steps for an iterative approach to print the Mountain Sequence Pattern for a given number N:

1. Run two nested loops.
2. The outer loop will care for the row of the pattern.
3. The inner loop will be caring for the column of the pattern.
4. Take three variable k1, k2, and gap which helps in generating a pattern.
5. After printing the row of the pattern update the value of k1 and k2 as: 
   • k1 = k1 + gap
   • k2 = k2 + gap

Below is the implementation of the iterative approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to create the mountain
// sequence pattern
void printPatt(int n)
{
    int k1 = 3;
    int k2 = 3;
    int gap = 5;

    // Outer loop to handle the row
    for (int i = 1; i <= 3; i++) {

        // Inner loop to handle the
        // Column
        for (int j = 1;
             j <= (5 * n); j++) {

            if (j > k2 && i < 3) {
                k2 += gap;
                k1 += gap;
            }

            // Condition to print the
            // star in mountain pattern
            if (j >= k1 && j <= k2) {
                cout << "*";
            }
            else {
                cout << " ";
            }
        }

        // Condition to adjust the value of
        // K1 and K2 for printing desire
        // Pattern
        if (i + 1 == 3) {
            k1 = 1;
            k2 = (5 * n);
        }
        else {
            k1 = 3;
            k2 = 3;
            k1--;
            k2++;
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    // Given Number N
    int N = 5;

    // Function call
    printPatt(N);
}

Java

// Java implementation of the above approach
class GFG{

// Function to create the mountain
// sequence pattern
static void printPatt(int n)
{
    int k1 = 3;
    int k2 = 3;
    int gap = 5;

    // Outer loop to handle the row
    for(int i = 1; i <= 3; i++)
    {

       // Inner loop to handle the
       // Column
       for(int j = 1; j <= (5 * n); j++)
       {
          if (j > k2 && i < 3)
          {
              k2 += gap;
              k1 += gap;
          }

          // Condition to print the
          // star in mountain pattern
          if (j >= k1 && j <= k2)
          {
              System.out.print("*");
          }
          else
          {
              System.out.print(" ");
          }
       }

       // Condition to adjust the value of
       // K1 and K2 for printing desire
       // Pattern
       if (i + 1 == 3)
       {
           k1 = 1;
           k2 = (5 * n);
       }
       else
       {
           k1 = 3;
           k2 = 3;
           k1--;
           k2++;
       }
       System.out.println();
    }
}

// Driver code
public static void main (String[] args)
{

    // Given Number N
    int N = 5;

    // Function call
    printPatt(N);
}
}

// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach

# Function to create the mountain
# sequence pattern
def printPatt(n):

    k1 = 3; k2 = 3; gap = 5;

    # Outer loop to handle the row
    for i in range(1, 4):

        # Inner loop to handle the
        # Column
        for j in range(1, (5 * n) + 1):

            if (j > k2 and i < 3):
                k2 += gap;
                k1 += gap;

            # Condition to print the
            # star in mountain pattern
            if (j >= k1 and j <= k2):
                print("*", end = "");
            else:
                print(" ", end = "");
        print("\n", end = "");

        # Condition to adjust the value of
        # K1 and K2 for printing desire
        # Pattern
        if (i + 1 == 3):
            k1 = 1;
            k2 = (5 * n);

        else:
            k1 = 3;
            k2 = 3;
            k1 -= 1;
            k2 += 1;
    print(end = "");

# Driver Code

# Given Number N
N = 5;

# Function call
printPatt(N);

# This code is contributed by Code_Mech

C#

// C# implementation of the above approach
using System;
class GFG{

// Function to create the mountain
// sequence pattern
static void printPatt(int n)
{
    int k1 = 3;
    int k2 = 3;
    int gap = 5;

    // Outer loop to handle the row
    for(int i = 1; i <= 3; i++)
    {

        // Inner loop to handle the
        // Column
        for(int j = 1; j <= (5 * n); j++)
        {
            if (j > k2 && i < 3)
            {
                k2 += gap;
                k1 += gap;
            }

            // Condition to print the
            // star in mountain pattern
            if (j >= k1 && j <= k2)
            {
                Console.Write("*");
            }
            else
            {
                Console.Write(" ");
            }
        }

        // Condition to adjust the value of
        // K1 and K2 for printing desire
        // Pattern
        if (i + 1 == 3)
        {
            k1 = 1;
            k2 = (5 * n);
        }
        else
        {
            k1 = 3;
            k2 = 3;
            k1--;
            k2++;
        }
        Console.WriteLine();
    }
}

// Driver code
public static void Main (String[] args)
{

    // Given Number N
    int N = 5;

    // Function call
    printPatt(N);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<!--  Javascript program for the above approach. -->

<script>

// Function to create the mountain
// sequence pattern
function printPatt( n)
{
    var k1 = 3;
    var k2 = 3;
    var gap = 5;

    // Outer loop to handle the row
    for(let i = 1; i <= 3; i++)
    {

        // Inner loop to handle the
        // Column
        for(let j = 1; j <= (5 * n); j++)
        {
            if (j > k2 && i < 3)
            {
                k2 += gap;
                k1 += gap;
            }

            // Condition to print the
            // star in mountain pattern
            if (j >= k1 && j <= k2)
            {   document.write("*");
            }
            else
            {
                document.write("  ");
            }
        }

        // Condition to adjust the value of
        // K1 and K2 for printing desire
        // Pattern
        if (i + 1 == 3)
        {
            k1 = 1;
            k2 = (5 * n);
        }
        else
        {
            k1 = 3;
            k2 = 3;
            k1--;
            k2++;
        }
        document.write("<br>");
    }
}

//Driver Code
var N=3;
printPatt(N);

</script>

<!--  This code in contributed by nirajgusain5 -->

Output

  *    *    *    *    *  
 ***  ***  ***  ***  *** 
*************************

Time Complexity: O(N)

Auxiliary Space: O(1)
Recursive Approach: The pattern can be generated using Recursion. Below are the steps:

1. Run two nested loops.
2. The outer loop will care for the row of the pattern.
3. The inner loop will be caring for the column of the pattern.
4. Apart from these, variables K1, K2, and gap are needed.
5. K1 and K2 will cover the cases when the * is to be printed.
6. The gap will cover the cases when spaces are to be printed.
7. Recursively call the function fun(i, j + 1) for handling columns.
8. Recursive call the function fun(i + 1, 0) for handling rows.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;
int k1 = 2;
int k2 = 2;
int gap = 5;

// Function to print pattern
// recursively
int printPattern(
    int i, int j, int n)
{

    // Base Case
    if (j >= n) {
        k1 = 2;
        k2 = 2;
        k1--;
        k2++;
        if (i == 2) {
            k1 = 0;
            k2 = n - 1;
        }
        return 0;
    }

    // Condition to check row limit
    if (i >= 3) {
        return 1;
    }

    // Condition for assigning gaps
    if (j > k2) {
        k1 += gap;
        k2 += gap;
    }

    // Conditions to print *
    if (j >= k1
            && j <= k2
        || i == 2) {

        cout << "*";
    }

    // Else print ' '
    else {

        cout << " ";
    }

    // Recursive call for columns
    if (printPattern(i, j + 1, n)
        == 1) {
        return 1;
    }

    cout << endl;

    // Recursive call for rows
    return printPattern(i + 1,
                        0, n);
}

// Driver Code
int main()
{
    // Given Number N
    int N = 3;

    // Function Call
    printPattern(0, 0, N * 5);
    return 0;
}

Java

import java.io.*;

// Java program for the
// above approach
class GFG {

    static int k1 = 2;
    static int k2 = 2;
    static int gap = 5;

    // Function to print pattern
    // recursively
    public static int printPattern(int i, int j, int n)
    {
        // Base Case
        if (j >= n) {
            k1 = 2;
            k2 = 2;
            k1--;
            k2++;
            if (i == 2) {
                k1 = 0;
                k2 = n - 1;
            }
            return 0;
        }

        // Condition to check
        // row limit
        if (i >= 3) {
            return 1;
        }

        // Condition for assigning gaps
        if (j > k2) {
            k1 += gap;
            k2 += gap;
        }

        // Conditions to print *
        if (j >= k1 && j <= k2 || i == 2) {
            System.out.print("*");
        }

        // Else print ' '
        else {
            System.out.print(" ");
        }

        // Recursive call for columns
        if (printPattern(i, j + 1, n) == 1) {
            return 1;
        }

        System.out.println();

        // Recursive call for rows
        return printPattern(i + 1, 0, n);
    }

    // Driver code
    public static void main(String[] args)
    {
        // Given Number N
        int N = 3;

        // Function Call
        printPattern(0, 0, N * 5);
    }
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the
# above approach
k1 = 2
k2 = 2
gap = 5

# Function to print pattern
# recursively
def printPattern(i, j, n):

    global k1
    global k2
    global gap

    # Base Case
    if(j >= n):
        k1 = 2
        k2 = 2
        k1 -= 1
        k2 += 1
        if(i == 2):
            k1 = 0
            k2 = n - 1
        return 0

    # Condition to check row limit
    if(i >= 3):
        return 1

    # Condition for assigning gaps
    if(j > k2):
        k1 += gap
        k2 += gap

    # Conditions to print *
    if(j >= k1 and j <= k2 or
       i == 2):
        print("*", end = "")

    # Else print ' '
    else:
        print(" ", end = "")

    # Recursive call for columns
    if(printPattern(i, j + 1, n) == 1):
        return 1

    print()

    # Recursive call for rows
    return (printPattern(i + 1, 0, n))

# Driver Code

# Given Number N
N = 3

# Function Call
printPattern(0, 0, N * 5)

#This code is contributed by avanitrachhadiya2155

C#

// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG {

    static int k1 = 2;
    static int k2 = 2;
    static int gap = 5;

    // Function to print pattern
    // recursively
    static int printPattern(int i, int j, int n)
    {
      // Base Case
      if (j >= n)
      {
        k1 = 2;
        k2 = 2;
        k1--;
        k2++;
        if (i == 2)
        {
          k1 = 0;
          k2 = n - 1;
        }
        return 0;
      }

      // Condition to check
      // row limit
      if (i >= 3)
      {
        return 1;
      }

      // Condition for assigning gaps
      if (j > k2)
      {
        k1 += gap;
        k2 += gap;
      }

      // Conditions to print *
      if (j >= k1 && j <= k2 || i == 2)
      {
        Console.Write("*");
      }

      // Else print ' '
      else
      {
        Console.Write(" ");
      }

      // Recursive call for columns
      if (printPattern(i, j + 1, n) == 1)
      {
        return 1;
      }

      Console.WriteLine();

      // Recursive call for rows
      return printPattern(i + 1, 0, n);
    }

  // Driver code
  static void Main()
  {

      // Given Number N
      int N = 3;

      // Function Call
      printPattern(0, 0, N * 5);
  }
}

// This code is contributed by divyeshrabadiya07

Javascript

// Javascript program for the above approach
let k1 = 2;
let k2 = 2;
let gap = 5;

// Function to print pattern
// recursively
function printPattern(i, j, n)
{
    // Base Case
    if (j >= n) {
        k1 = 2;
        k2 = 2;
        k1--;
        k2++;
        if (i == 2) {
            k1 = 0;
            k2 = n - 1;
        }
            return 0;
        }

    // Condition to check row limit
    if (i >= 3) {
        return 1;
    }

    // Condition for assigning gaps
    if (j > k2) {
        k1 += gap;
        k2 += gap;
    }

    // Conditions to print *
    if (j >= k1    && j <= k2 || i == 2) {

        console.log("*");
    }

    // Else print ' '
    else {
      console.log("\xa0\xa0");
    }

    // Recursive call for columns
    if (printPattern(i, j + 1, n) == 1) {
        return 1;
    }

    console.log("<br>");

    // Recursive call for rows
    return printPattern(i + 1, 0, n);
}

// Driver Code

    // Given Number N
    let N = 3;

    // Function Call
    printPattern(0, 0, N * 5);

    // This code is contributed by Pushpesh Raj.

Output

  *    *    *  
 ***  ***  *** 
***************

Time Complexity: O(N)
Auxiliary Space: O(N) for call stack