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:
- Run two nested loops.
- The outer loop will care for the row of the pattern.
- The inner loop will be caring for the column of the pattern.
- Take three variable k1, k2, and gap which helps in generating a pattern.
- 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 patternvoid 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 Codeint 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 patternstatic 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 patterndef 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 NN = 5;# Function callprintPatt(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 patternstatic 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 patternfunction 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 Codevar 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:
- Run two nested loops.
- The outer loop will care for the row of the pattern.
- The inner loop will be caring for the column of the pattern.
- Apart from these, variables K1, K2, and gap are needed.
- K1 and K2 will cover the cases when the * is to be printed.
- The gap will cover the cases when spaces are to be printed.
- Recursively call the function fun(i, j + 1) for handling columns.
- 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// recursivelyint 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 Codeint 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 approachclass 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 = 2k2 = 2gap = 5# Function to print pattern # recursivelydef 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 NN = 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// recursivelyfunction 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
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
