# 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:
*    *    *    *
***  ***  ***  ***
********************
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program for the above approach ` `#include ` `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); ` `} `

## 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 ` `       ``// 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  `

## 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 ` `        ``# 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 `

## 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 ` `        ``// 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  `

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++

 `// C++ program for the above approach ` ` `  `#include ` `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; ` `} `

Output:

```  *    *    *
***  ***  ***
***************
```

Time Complexity: O(N)

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