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 iterative approach to print the Mountain Sequence Pattern for a given number N:

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

Below is the implementation of iterative approach:

C++

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// 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
        // Coloumn
        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);
}

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Java

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// 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
       // Coloumn
       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 

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Python3

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# 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
        # Coloumn
        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

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C#

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// 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
        // Coloumn
        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 

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Output:



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

Time Complexity: O(N)

Recursive Approach: The pattern can be generated using Recursion. Below are the steps:

  1. Run two nested loop.
  2. outer loop will care the row of the pattern.
  3. Inner loop will be caring the column of the pattern.
  4. Apart from these, variables K1, K2 and gap are needed.
  5. K1, K2 will cover the cases when the * is to be printed.
  6. 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 above approach:

C++

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// 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;
}

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Output:

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

Time Complexity: O(N)

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